Secant and Tangent Lines

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Secant and Tangent Lines

http://www.mathscoop.com/applets/tangent-line-applet.php f (a + h) − f (a) h

Slope of secant line:

Slope of tangent line: f (a) f (a + h) f (a + h) − f (a) lim h→0 h

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y y = f (x)

x a a + h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1 f (a + h) − f (a) h

Slope of secant line:

Slope of tangent line: f (a) f (a + h) f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y y = f (x)

x a a + h f (a + h) − f (a) h

Slope of secant line:

Slope of tangent line: f (a) f (a + h) f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y y = f (x)

x a a + h f (a + h) − f (a) h

Slope of secant line:

Slope of tangent line: f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y y = f (x)

f (a) f (a + h)

x a a + h f (a + h) − f (a) h

Slope of tangent line: f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y Slope of secant line: y = f (x)

f (a) f (a + h)

x a a + h h

Slope of tangent line: f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y Slope of secant line: y = f (x) f (a + h) − f (a)

f (a) f (a + h)

x a a + h Slope of tangent line: f (a + h) − f (a) lim h→0 h

Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y Slope of secant line: y = f (x) f (a + h) − f (a) h

f (a) f (a + h)

x a a + h Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y Slope of secant line: y = f (x) f (a + h) − f (a) h

Slope of tangent line: f (a) f (a + h) f (a + h) − f (a) lim h→0 h x a a + h Chapter 2: Derivatives 2.1: Revisiting Tangent Lines Slope of Secant and Tangent Line

Slope Recall, the slope of a line is given by any of the following:

rise ∆y y2 − y1 run ∆x x2 − x1

Suppose we’re interested in the slope of the secant line to f (x) through the points where x = a and x = a + h.

y Slope of secant line: y = f (x) f (a + h) − f (a) h

Slope of tangent line: f (a) f (a + h) f (a + h) − f (a) lim h→0 h x a a + h increasing It “points up.” decreasing It “points down.”

If f 0(a) > 0, then f is at a. If f 0(a) < 0, then f is at a. If f 0(a) = 0, then f looks constant at a. It looks ﬂat.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a. decreasing It “points down.”

increasing It “points up.” If f 0(a) < 0, then f is at a. If f 0(a) = 0, then f looks constant at a. It looks ﬂat.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a.

If f 0(a) > 0, then f is at a. decreasing It “points down.”

If f 0(a) < 0, then f is at a. If f 0(a) = 0, then f looks constant at a. It looks ﬂat.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a.

If f 0(a) > 0, then f is increasing at a. It “points up.” decreasing It “points down.” If f 0(a) = 0, then f looks constant at a. It looks ﬂat.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a.

If f 0(a) > 0, then f is increasing at a. It “points up.” If f 0(a) < 0, then f is at a. If f 0(a) = 0, then f looks constant at a. It looks ﬂat.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a.

If f 0(a) > 0, then f is increasing at a. It “points up.” If f 0(a) < 0, then f is decreasing at a. It “points down.” Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivative at a Point

Deﬁnition Given a function f (x) and a point a, the slope of the tangent line to f (x) at a is the derivative of f at a, written f 0(a).

f (a + h) − f (a) So, f 0(a) = lim . h→0 h

f 0(a) is also the instantaneous rate of change of f at a.

If f 0(a) > 0, then f is increasing at a. It “points up.” If f 0(a) < 0, then f is decreasing at a. It “points down.” If f 0(a) = 0, then f looks constant at a. It looks ﬂat. a b c d

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) < 0? Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) < 0? f 0(b) < 0 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) > 0? Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) > 0? f 0(a) > 0 and f 0(d) > 0 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) = 0? Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Practice: Increasing and Decreasing

x a b c d

Where is f 0(x) = 0? f 0(c) = 0 f (3 + h) − f (3) f 0(3) = lim h→0 h f (3 + h) (3 + h)2 − 5 − (32 − 5) = lim h→0 h 9 + 6h + h2 − 5 − 4 = lim h→0 h h2 + 6h = lim h→0 h = lim h + 6 h→0 f (3) = 6

3 + h

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y

x 3 f (3 + h) (3 + h)2 − 5 − (32 − 5) = lim h→0 h 9 + 6h + h2 − 5 − 4 = lim h→0 h h2 + 6h = lim h→0 h = lim h + 6 h→0 f (3) = 6

3 + h

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y f (3 + h) − f (3) f 0(3) = lim h→0 h

x 3 f (3 + h) (3 + h)2 − 5 − (32 − 5) = lim h→0 h 9 + 6h + h2 − 5 − 4 = lim h→0 h h2 + 6h = lim h→0 h = lim h + 6 h→0 f (3) = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y f (3 + h) − f (3) f 0(3) = lim h→0 h

x 3 3 + h (3 + h)2 − 5 − (32 − 5) = lim h→0 h 9 + 6h + h2 − 5 − 4 = lim h→0 h h2 + 6h = lim h→0 h = lim h + 6 h→0 = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y f (3 + h) − f (3) f 0(3) = lim h→0 h f (3 + h)

f (3)

x 3 3 + h 9 + 6h + h2 − 5 − 4 = lim h→0 h h2 + 6h = lim h→0 h = lim h + 6 h→0 = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y f (3 + h) − f (3) f 0(3) = lim h→0 h f (3 + h) (3 + h)2 − 5 − (32 − 5) = lim h→0 h

f (3)

x 3 3 + h h2 + 6h = lim h→0 h = lim h + 6 h→0 = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Use the deﬁnition of the derivative to ﬁnd the slope of the tangent line to f (x) = x 2 − 5 at the point x = 3. y f (3 + h) − f (3) f 0(3) = lim h→0 h f (3 + h) (3 + h)2 − 5 − (32 − 5) = lim h→0 h 9 + 6h + h2 − 5 − 4 = lim h→0 h

f (3)

x 3 3 + h = lim h + 6 h→0 = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

f (3)

x 3 3 + h = 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

x 3 3 + h Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

x 3 3 + h f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h=2x In particular, f 0(3) = 6. h→0

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x: (x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h=2x In particular, f 0(3) = 6. h→0

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h=2x In particular, f 0(3) = 6. h→0

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h 2xh + h2 = lim h→0 h

= lim 2x + h=2x In particular, f 0(3) = 6. h→0

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h = lim 2x + h=2x In particular, f 0(3) = 6. h→0

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h =2x In particular, f 0(3) = 6.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h h→0 In particular, f 0(3) = 6.

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h=2x h→0 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative The Derivative as a Function

Let’s keep the function f (x) = x 2 − 5. We just showed f 0(3) = 6. We can also ﬁnd its derivative at an arbitrary point x:

f (x + h) − f (x) f 0(x) = lim h→0 h

(x + h)2 − 5 − (x 2 − 5) = lim h→0 h

x 2 + 2xh + h2 − 5 − x 2 + 5 = lim h→0 h

2xh + h2 = lim h→0 h

= lim 2x + h=2x In particular, f 0(3) = 6. h→0 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

0 y f (x) = 2x

f (x) = x 2 − 5

x Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Increasing and Decreasing

In black is the curve y = f (x). Which of the coloured curves corresponds to y = f 0(x)? y y = f (x)

y y y

x x x

A B C Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Increasing and Decreasing

In black is the curve y = f (x). Which of the coloured curves corresponds to y = f 0(x)?

y y = f (x)

y y y

x x x

A B C Notation The ”prime” notation f 0(x) and f 0(a) is attributed to Newton. We will also use Leibnitz notation:

df df d d (a) f (x) f (x) dx dx dx dx x=a function number function number

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Derivatives Let f (x) be a function. The derivative of f (x) with respect to x is given by

f (x + h) − f (x) f 0(x) = lim , h→0 h provided the limit exists. Notice that x will be a part of your ﬁnal expression: this is a function.

If f 0(x) exists for all x in an interval (a, b), we say that f is diﬀerentiable on (a, b). function number function number

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Derivatives Let f (x) be a function. The derivative of f (x) with respect to x is given by

f (x + h) − f (x) f 0(x) = lim , h→0 h provided the limit exists. Notice that x will be a part of your ﬁnal expression: this is a function.

If f 0(x) exists for all x in an interval (a, b), we say that f is diﬀerentiable on (a, b).

Notation The ”prime” notation f 0(x) and f 0(a) is attributed to Newton. We will also use Leibnitz notation:

df df d d (a) f (x) f (x) dx dx dx dx x=a Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Derivatives Let f (x) be a function. The derivative of f (x) with respect to x is given by

f (x + h) − f (x) f 0(x) = lim , h→0 h provided the limit exists. Notice that x will be a part of your ﬁnal expression: this is a function.

If f 0(x) exists for all x in an interval (a, b), we say that f is diﬀerentiable on (a, b).

Notation The ”prime” notation f 0(x) and f 0(a) is attributed to Newton. We will also use Leibnitz notation:

df df d d (a) f (x) f (x) dx dx dx dx x=a function number function number 2x 6 2x 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Alternate Notation

Newtonian Notation: f (x) = x 2 + 5 f 0(x) = 2x f 0(3) = 6 Leibnitz Notation: df = df (3) = d f (x) = d f (x) = dx dx dx dx x=3 y f 0(x) = 2x f (x) = x 2 − 5

x 6 2x 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Alternate Notation

Newtonian Notation: f (x) = x 2 + 5 f 0(x) = 2x f 0(3) = 6 Leibnitz Notation: df = 2x df (3) = d f (x) = d f (x) = dx dx dx dx x=3 y f 0(x) = 2x f (x) = x 2 − 5

x 2x 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Alternate Notation

Newtonian Notation: f (x) = x 2 + 5 f 0(x) = 2x f 0(3) = 6 Leibnitz Notation: df = 2x df (3) = 6 d f (x) = d f (x) = dx dx dx dx x=3 y f 0(x) = 2x f (x) = x 2 − 5

x 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Alternate Notation

Newtonian Notation: f (x) = x 2 + 5 f 0(x) = 2x f 0(3) = 6 Leibnitz Notation: df = 2x df (3) = 6 d f (x) = 2x d f (x) = dx dx dx dx x=3 y f 0(x) = 2x f (x) = x 2 − 5

x Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Alternate Notation

Newtonian Notation: f (x) = x 2 + 5 f 0(x) = 2x f 0(3) = 6 Leibnitz Notation: df = 2x df (3) = 6 d f (x) = 2x d f (x) = 6 dx dx dx dx x=3 y f 0(x) = 2x f (x) = x 2 − 5

x Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Alternate Deﬁnition Calculating f (a + h) − f (a) f 0(a) = lim h→0 h is the same as calculating f (x) − f (a) f 0(x) = lim . x→a x − a Notice in these scenarios, h = x − a.

a x f (x + h) − f (x) f 0(x) = lim √h→0 √ h x + h − x = lim h→0 √ h √ √ √ x + h − x x + h + x = lim √ √ h→0 h x + h + x (x + h) − (x) = lim √ √ h→0 h( x + h + x) 1 = lim √ √ h→0 x + h + x 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

√ Let f (x) = x. Using the deﬁnition of a derivative, calculate f 0(x). √ √ x + h − x = lim h→0 √ h √ √ √ x + h − x x + h + x = lim √ √ h→0 h x + h + x (x + h) − (x) = lim √ √ h→0 h( x + h + x) 1 = lim √ √ h→0 x + h + x 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

√ Let f (x) = x. Using the deﬁnition of a derivative, calculate f 0(x). f (x + h) − f (x) f 0(x) = lim h→0 h √ √ √ √ x + h − x x + h + x = lim √ √ h→0 h x + h + x (x + h) − (x) = lim √ √ h→0 h( x + h + x) 1 = lim √ √ h→0 x + h + x 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

√ Let f (x) = x. Using the deﬁnition of a derivative, calculate f 0(x). f (x + h) − f (x) f 0(x) = lim √h→0 √ h x + h − x = lim h→0 h (x + h) − (x) = lim √ √ h→0 h( x + h + x) 1 = lim √ √ h→0 x + h + x 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

√ Let f (x) = x. Using the deﬁnition of a derivative, calculate f 0(x). f (x + h) − f (x) f 0(x) = lim √h→0 √ h x + h − x = lim h→0 √ h √ √ √ x + h − x x + h + x = lim √ √ h→0 h x + h + x 1 = lim √ √ h→0 x + h + x 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

√ Let f (x) = x. Using the deﬁnition of a derivative, calculate f 0(x). f (x + h) − f (x) f 0(x) = lim √h→0 √ h x + h − x = lim h→0 √ h √ √ √ x + h − x x + h + x = lim √ √ h→0 h x + h + x (x + h) − (x) = lim √ √ h→0 h( x + h + x) 1 = √ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of a Derivative

∞ 0 √ 1 lim x = lim √ = x→0+ x→0+ 2 x

√ 1 Review: lim x = lim √ = x→∞ x→∞ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative y

√ y = x

√1 y = 2 x x 0 ∞

∞ 0 √ 1 lim x = lim √ = x→0+ x→0+ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = lim √ = x→∞ x→∞ 2 x 0 ∞

0 √ 1 lim x = lim √ = x→0+ x→0+ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = ∞ lim √ = x→∞ x→∞ 2 x 0 ∞

√ 1 lim x = lim √ = x→0+ x→0+ 2 x

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = ∞ lim √ = 0 x→∞ x→∞ 2 x 0 ∞

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = ∞ lim √ = 0 x→∞ x→∞ 2 x √ 1 lim x = lim √ = x→0+ x→0+ 2 x ∞

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = ∞ lim √ = 0 x→∞ x→∞ 2 x √ 1 lim x = 0 lim √ = x→0+ x→0+ 2 x Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

√ y = x

√1 y = 2 x x

√ 1 Review: lim x = ∞ lim √ = 0 x→∞ x→∞ 2 x √ 1 lim x = 0 lim √ = ∞ x→0+ x→0+ 2 x x − x+h = lim x(x+h) x(x+h) h→0 h −h = lim x(x+h) h→0 h −1 = lim h→0 x(x + h) 1 = − x 2

d 1 1 − 1 = lim x+h x dx x h→0 h

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of the Derivative

d 1 Using the deﬁnition of the derivative, calculate . dx x x − x+h = lim x(x+h) x(x+h) h→0 h −h = lim x(x+h) h→0 h −1 = lim h→0 x(x + h) 1 = − x 2

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of the Derivative

d 1 Using the deﬁnition of the derivative, calculate . dx x

d 1 1 − 1 = lim x+h x dx x h→0 h Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of the Derivative

d 1 Using the deﬁnition of the derivative, calculate . dx x

d 1 1 − 1 = lim x+h x dx x h→0 h x − x+h = lim x(x+h) x(x+h) h→0 h −h = lim x(x+h) h→0 h −1 = lim h→0 x(x + h) 1 = − x 2 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Using the Deﬁnition of the Derivative

d 2x Using the deﬁnition of the derivative, calculate . dx x + 1

d 1 Using the definition of the derivative, calculate √ . dx x 2 + x Chapter 2: Derivatives 2.2: Definition of the Derivative Using the Definition of the Derivative

d 2x Using the deﬁnition of the derivative, calculate . dx x + 1

2(x+h) f (x + h) − f (x) − 2x 1 lim = lim x+h+1 x+1 = lim 2(x+h)(x+1) − 2x(x+h+1) h→0 h h→0 h h→0 h (x+h+1)(x+1) (x+1)(x+h+1) 2 (x 2 + x + xh + h) − (x 2 + xh + x) = lim h→0 h (x + h + 1)(x + 1) 2 h = lim h→0 h (x + h + 1)(x + 1) 2 2 = lim = h→0 (x + h + 1)(x + 1) (x + 1)2

d 1 Using the definition of the derivative, calculate √ . dx x 2 + x Chapter 2: Derivatives 2.2: Definition of the Derivative Using the Definition of the Derivative

d 2x Using the deﬁnition of the derivative, calculate . dx x + 1 d 1 Using the deﬁnition of the derivative, calculate √ . dx x 2 + x

√ 1 − √ 1 f (x + h) − f (x) (x+h)2+x+h x2+x lim = lim h→0 h h→0 h √ √ 1 2 2 = lim √ x +x √ − √ (x+h) +x√+h h→0 h (x2+h)2+x+h x2+x (x2+h)2+x+h x2+x √ √ √ √ 1 2 2 2 2 = lim √x +x− (x+h√) +x+h √x +x+√(x+h) +x+h h→0 h (x2+h)2+x+h x2+x x2+x+ (x+h)2+x+h 1 (x2+x)−[(x+h)2+x+h] = lim √ √ h√ √ i h→0 h (x2+h)2+x+h x2+x x2+x+ (x+h)2+x+h 1 −(2xh+h2+h) = lim √ √ h√ √ i h→0 h (x2+h)2+x+h x2+x x2+x+ (x+h)2+x+h −(2x+h+1) −(2x+1) = lim √ √ h√ √ i = 2 3/2 h→0 (x2+h)2+x+h x2+x x2+x+ (x+h)2+x+h 2(x +x) Rearranging: y − f (a) = f 0(a)(x − a) ← equation of tangent line

(x, y)

Slope of the tangent line: rise y − f (a) f 0(a) = = run x − a

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

The tangent line to f (x) at a has slope f 0(a) and passes through the point (a, f (a)). y

y = f (x) (a, f (a))

x Rearranging: y − f (a) = f 0(a)(x − a) ← equation of tangent line

Slope of the tangent line: rise y − f (a) f 0(a) = = run x − a

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

The tangent line to f (x) at a has slope f 0(a) and passes through the point (a, f (a)). y (x, y)

y = f (x) (a, f (a))

x Rearranging: y − f (a) = f 0(a)(x − a) ← equation of tangent line

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

The tangent line to f (x) at a has slope f 0(a) and passes through the point (a, f (a)). y (x, y)

y = f (x) (a, f (a))

Slope of the tangent line: rise y − f (a) f 0(a) = = run x − a ← equation of tangent line

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

The tangent line to f (x) at a has slope f 0(a) and passes through the point (a, f (a)). y (x, y)

y = f (x) (a, f (a))

Slope of the tangent line: rise y − f (a) f 0(a) = = run x − a Rearranging: y − f (a) = f 0(a)(x − a) Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

The tangent line to f (x) at a has slope f 0(a) and passes through the point (a, f (a)). y (x, y)

y = f (x) (a, f (a))

Slope of the tangent line: rise y − f (a) f 0(a) = = run x − a Rearranging: y − f (a) = f 0(a)(x − a) ← equation of tangent line 9 3 √1 = 1 = 1 2 9 2(3) 6

a = f (a) = f 0(a) = 1 (y − 3) = (x − 9) 6

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x .

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a) 9 3 √1 = 1 = 1 2 9 2(3) 6

a = f (a) = f 0(a) = 1 (y − 3) = (x − 9) 6

√ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x .

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) 9 3 √1 = 1 = 1 2 9 2(3) 6

a = f (a) = f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . 3 √1 = 1 = 1 2 9 2(3) 6

9 f (a) = f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a = 3 √1 = 1 = 1 2 9 2(3) 6

f (a) = f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 √1 = 1 = 1 2 9 2(3) 6

3 f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 f (a) = √1 = 1 = 1 2 9 2(3) 6

f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 f (a) = 3 √1 = 1 = 1 2 9 2(3) 6 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 f (a) = 3 f 0(a) = 1 (y − 3) = (x − 9) 6

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 f (a) = 3 f 0(a) = √1 = 1 = 1 2 9 2(3) 6 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Equation of the Tangent Line

Point-Slope Formula The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

In general, a line with slope m passing through point (x1, y1) has the equation:

(y − y1) = m(x − x1) √ Find the equation√ of the tangent line to the curve f (x) = x at x = 9. d √1 Recall dx x = 2 x . a =9 f (a) = 3 f 0(a) = √1 = 1 = 1 2 9 2(3) 6 1 (y − 3) = (x − 9) 6 1 2.2 −1.6

(y − 2.2) = −1.6(x − 1)

a = s(a) = s0(a) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

Things to have Memorized The derivative of a function f at a point a is given by the following limit, if it exists: f (a + h) − f (a) f 0(a) = lim h→0 h The tangent line to the function f (x) at point a is:

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. 1 2.2 −1.6

(y − 2.2) = −1.6(x − 1)

s(a) = s0(a) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = 2.2 −1.6

(y − 2.2) = −1.6(x − 1)

s0(a) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = s(a) = 1 2.2 −1.6

(y − 2.2) = −1.6(x − 1)

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = s(a) = s0(a) = 2.2 −1.6

(y − 2.2) = −1.6(x − 1)

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = 1 s(a) = s0(a) = −1.6

(y − 2.2) = −1.6(x − 1)

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = 1 s(a) =2.2 s0(a) = (y − 2.2) = −1.6(x − 1)

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = 1 s(a) =2.2 s0(a) =−1.6 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

(y − f (a)) = f 0(a)(x − a)

Let s(t) = 3 − 0.8t2. Then s0(t) = −1.6t. Find the equation for the tangent line to the function s(t) when t = 1. a = 1 s(a) =2.2 s0(a) =−1.6

(y − 2.2) = −1.6(x − 1) A. 0 B. 1 C. 2 D. −15 E. −13

2x − 15

f 0(1) =

f 0(5) =2

f 0(−13) =2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is: A. 0 B. 1 C. 2 D. −15 E. −13

f 0(1) =

f 0(5) =2

f 0(−13) =2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(5) =2

f 0(−13) =2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C. 2 D. −15 E. −13 2

f 0(−13) =2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) = f 0(−13) =2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 2

g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) = g(x) = 13

g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) =2 g 0(1) =

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) =2

g(x) = 13 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) =2

g(x) = 13

g 0(1) = Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) =2

g(x) = 13

g 0(1) = A. 0 B. 1 C. 2 D. 13 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Derivatives of Lines

f (x) = 2x − 15 The equation of the tangent line to f (x) at x = 100 is:

2x − 15 f 0(1) = A. 0 B. 1 C.2D. −15 E. −13 f 0(5) =2 f 0(−13) =2

g(x) = 13

g 0(1) =A.0 B. 1 C. 2 D. 13 f (x) + c

Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

f (x)

x Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

x Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

x Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

x Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

x Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

x d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

Adding or subtracting a constant to a function does not change its derivative. ⇒ d 10 − 0.8t2 = −1.6 dx t=1

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 dx t=1 Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Adding a Constant

y f (x) + c

f (x)

Adding or subtracting a constant to a function does not change its derivative.

d 3 − 0.8t2 = −1.6 ⇒ d 10 − 0.8t2 = −1.6 dx t=1 dx t=1 ← Deriv of constant is zero

←Add a constant: no change

← Multiply by a constant: keep the constant

Derivatives From Previous Calculations d dx {2x − 15} = 2 d x 2 − 5 = 2x dx √ d √1 dx x = 2 x d dx {13} = 0 √ 0 √1 For instance: let f (x) = 10 (2x − 15)+ 13 − x . Then f (x)= 10 (2)+0+ 2 x .

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x) ←Add a constant: no change

← Multiply by a constant: keep the constant

← Deriv of constant is zero √ 0 √1 For instance: let f (x) = 10 (2x − 15)+ 13 − x . Then f (x)= 10 (2)+0+ 2 x .

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x)

Derivatives From Previous Calculations d dx {2x − 15} = 2 d x 2 − 5 = 2x dx √ d √1 dx x = 2 x d dx {13} = 0 ←Add a constant: no change

← Multiply by a constant: keep the constant

← Deriv of constant is zero √1 = 10 (2)+0+ 2 x .

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x)

Derivatives From Previous Calculations d dx {2x − 15} = 2 d x 2 − 5 = 2x dx √ d √1 dx x = 2 x d dx {13} = 0 √ For instance: let f (x) = 10 (2x − 15)+ 13 − x. Then f 0(x) ←Add a constant: no change

← Multiply by a constant: keep the constant

← Deriv of constant is zero

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x)

Derivatives From Previous Calculations d dx {2x − 15} = 2 d x 2 − 5 = 2x dx √ d √1 dx x = 2 x d dx {13} = 0 √ 0 √1 For instance: let f (x) = 10 (2x − 15)+ 13 − x . Then f (x)= 10 (2)+0+ 2 x . ←Add a constant: no change

← Multiply by a constant: keep the constant

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x)

Derivatives From Previous Calculations d dx {2x − 15} = 2 d x 2 − 5 = 2x dx √ d √1 dx x = 2 x d dx {13} = 0 ← Deriv of constant is zero √ 0 √1 For instance: let f (x) = 10 (2x − 15)+ 13 − x . Then f (x)= 10 (2)+0+ 2 x . ← Multiply by a constant: keep the constant

Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative

Rules Suppose f (x) and g(x) are diﬀerentiable, and let c be a constant number. Then: d 0 0 dx {f (x) + g(x)} = f (x) + g (x) ←Add a constant: no change d 0 0 dx {f (x) − g(x)} = f (x) − g (x) d 0 dx {cf (x)} = cf (x)

Suppose f 0(x) = 3x, g 0(x) = −x 2, and h0(x) = 5. Calculate: d {f (x) + 5g(x) − h(x) + 22} dx A.3 x − 5x 2 B.3 x − 5x 2 − 5 C.3 x − 5x 2 − 5 + 22 D. none of the above Chapter 2: Derivatives 2.2: Deﬁnition of the Derivative Now You

Suppose f 0(x) = 3x, g 0(x) = −x 2, and h0(x) = 5. Calculate: d {f (x) + 5g(x) − h(x) + 22} dx A.3 x − 5x 2 B.3 x − 5x 2 − 5 C.3 x − 5x 2 − 5 + 22 D. none of the above Suppose P(t) gives the number of people in the world at t minutes past midnight, January 1, 2012. Suppose further that P0(0) = 156. How do you interpret P0(0) = 156? At midnight of January 1, 2012, the world population was increasing at a rate of 156 people each minute

Suppose P(n) gives the total profit, in dollars, earned by selling n widgets. How do you interpret P0(100)? How fast your profit is increasing as you sell more widgets, measured in dollars per widget, at the time you sell Widget #100. So, roughly the profit earned from the sale of the 100th widget.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

The derivative of f (x) at a, written f 0(a), is the instantaneous rate of change of f (x) when x = a. At midnight of January 1, 2012, the world population was increasing at a rate of 156 people each minute

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

The derivative of f (x) at a, written f 0(a), is the instantaneous rate of change of f (x) when x = a.

Suppose P(t) gives the number of people in the world at t minutes past midnight, January 1, 2012. Suppose further that P0(0) = 156. How do you interpret P0(0) = 156? Suppose P(n) gives the total profit, in dollars, earned by selling n widgets. How do you interpret P0(100)? How fast your profit is increasing as you sell more widgets, measured in dollars per widget, at the time you sell Widget #100. So, roughly the profit earned from the sale of the 100th widget.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

The derivative of f (x) at a, written f 0(a), is the instantaneous rate of change of f (x) when x = a.

Suppose P(t) gives the number of people in the world at t minutes past midnight, January 1, 2012. Suppose further that P0(0) = 156. How do you interpret P0(0) = 156? At midnight of January 1, 2012, the world population was increasing at a rate of 156 people each minute How fast your proﬁt is increasing as you sell more widgets, measured in dollars per widget, at the time you sell Widget #100. So, roughly the proﬁt earned from the sale of the 100th widget.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

The derivative of f (x) at a, written f 0(a), is the instantaneous rate of change of f (x) when x = a.

Suppose P(n) gives the total proﬁt, in dollars, earned by selling n widgets. How do you interpret P0(100)? Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

The derivative of f (x) at a, written f 0(a), is the instantaneous rate of change of f (x) when x = a.

The rate at which molecules of a certain type are being created or destroyed. Roughly, how many molecules are being added (or taken away, if negative) per second at time t.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

Suppose h(t) gives the height of a rocket t seconds after liftoﬀ. What is the interpretation of h0(t)?

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

Suppose h(t) gives the height of a rocket t seconds after liftoﬀ. What is the interpretation of h0(t)? The speed at which the rocket is rising at time t.

Suppose M(t) is the number of molecules of a chemical in a test tube t seconds after a reaction starts. Interpret M0(t). Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

Suppose h(t) gives the height of a rocket t seconds after liftoﬀ. What is the interpretation of h0(t)? The speed at which the rocket is rising at time t.

Suppose M(t) is the number of molecules of a chemical in a test tube t seconds after a reaction starts. Interpret M0(t). The rate at which molecules of a certain type are being created or destroyed. Roughly, how many molecules are being added (or taken away, if negative) per second at time t. When your load is about 100 kg, you need to increase the diameter of your wire by about 0.01 mm for each kg increase in your load. A paper1 on the impacts of various factors in average life expectancy contains the following: The only statistically significant variable in the [Least Developed Countries] model is physician density. The coefficient for this variable 20.67 indicating that a one unit increase in physician density leads to a 20.67 unit increase in life expectancy. This variable is also statistically significant at the 1% level demonstrating that this variable is very strongly and positively correlated with quality of healthcare received. This denotes that access to healthcare is very impactful in terms of increasing the quality of health in the country. If L(p) is the average life expectancy in an area with a density p of physicians, write the statement as a derivative: “a one unit increase in physician density leads to a 20.67 unit 0 increase in life expectancy.” L (p) = 20.67 Remark: physician density is measured as number of doctors per 1000 members of the population.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

Suppose G(w) gives the diameter in millimetres of steel wire needed to safely support a load of w kg. Suppose further that G 0(100) = 0.01. How do you interpret G 0(100) = 0.01?

1https://smartech.gatech.edu/bitstream/handle/1853/51648/The+Effect+of+National+ Healthcare+Expenditure+on+Life+Expectancy.pdf A paper1 on the impacts of various factors in average life expectancy contains the following: The only statistically significant variable in the [Least Developed Countries] model is physician density. The coefficient for this variable 20.67 indicating that a one unit increase in physician density leads to a 20.67 unit increase in life expectancy. This variable is also statistically significant at the 1% level demonstrating that this variable is very strongly and positively correlated with quality of healthcare received. This denotes that access to healthcare is very impactful in terms of increasing the quality of health in the country. If L(p) is the average life expectancy in an area with a density p of physicians, write the statement as a derivative: “a one unit increase in physician density leads to a 20.67 unit 0 increase in life expectancy.” L (p) = 20.67 Remark: physician density is measured as number of doctors per 1000 members of the population.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

1https://smartech.gatech.edu/bitstream/handle/1853/51648/The+Effect+of+National+ Healthcare+Expenditure+on+Life+Expectancy.pdf If L(p) is the average life expectancy in an area with a density p of physicians, write the statement as a derivative: “a one unit increase in physician density leads to a 20.67 unit 0 increase in life expectancy.” L (p) = 20.67 Remark: physician density is measured as number of doctors per 1000 members of the population.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

1https://smartech.gatech.edu/bitstream/handle/1853/51648/The+Effect+of+National+ Healthcare+Expenditure+on+Life+Expectancy.pdf 0 L (p) = 20.67 Remark: physician density is measured as number of doctors per 1000 members of the population.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

Suppose G(w) gives the diameter in millimetres of steel wire needed to safely support a load of w kg. Suppose further that G 0(100) = 0.01. How do you interpret G 0(100) = 0.01? When your load is about 100 kg, you need to increase the diameter of your wire by about 0.01 mm for each kg increase in your load. A paper1 on the impacts of various factors in average life expectancy contains the following: The only statistically significant variable in the [Least Developed Countries] model is physician density. The coefficient for this variable 20.67 indicating that a one unit increase in physician density leads to a 20.67 unit increase in life expectancy. This variable is also statistically significant at the 1% level demonstrating that this variable is very strongly and positively correlated with quality of healthcare received. This denotes that access to healthcare is very impactful in terms of increasing the quality of health in the country. If L(p) is the average life expectancy in an area with a density p of physicians, write the statement as a derivative: “a one unit increase in physician density leads to a 20.67 unit increase in life expectancy.”

1https://smartech.gatech.edu/bitstream/handle/1853/51648/The+Effect+of+National+ Healthcare+Expenditure+on+Life+Expectancy.pdf Remark: physician density is measured as number of doctors per 1000 members of the population.

Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

1https://smartech.gatech.edu/bitstream/handle/1853/51648/The+Effect+of+National+ Healthcare+Expenditure+on+Life+Expectancy.pdf Chapter 2: Derivatives 2.3: Interpreting the Derivative Interpreting the Derivative

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