Tangent Line, Velocity, Derivative and Differentiability

instant rate of change and approximation of the function

– Typeset by FoilTEX – 1 Motivation: what do tangent line and velocity have in common?

– Typeset by FoilTEX – 2 Euclid: The tangent is a line that touches a curve at just one point.

Works for circles: Euclid: The tangent is a line that touches a curve at just one point.

Works for circles:

x Does NOT work for sin(x)! x

– Typeset by FoilTEX – 3 Modern: The tangent is the limiting position of the secant line (if exists).

Tangent = lim Secant f(t) t→x slope of _f_(t_)-_f(_x_) = secant t-x Slope of Slope of = lim f(t)-f(x) t→x Tangent Secant f(x) t-x x t Slope of f(t) − f(x) = lim Tangent t→x t − x

– Typeset by FoilTEX – 4 EXAMPLES

– Typeset by FoilTEX – 5 Slope of f(t) − f(x) = lim Tangent t→x t − x

EXAMPLE 1. Find slope of the tangent line to the graph of f(x) = x2 − 4 at point (3, 5). Slope of f(t) − f(x) = lim Tangent t→x t − x

EXAMPLE 1. Find slope of the tangent line to the graph of f(x) = x2 − 4 at point (3, 5). Slope of (t2 − 4) − (x2 − 4) = lim Tangent t→x t − x Slope of f(t) − f(x) = lim Tangent t→x t − x

EXAMPLE 1. Find slope of the tangent line to the graph of f(x) = x2 − 4 at point (3, 5). Slope of (t2 − 4) − (x2 − 4) = lim Tangent t→x t − x t2 − x2 (t − x)(t + x) = lim = lim = 2x t→x t − x t→x t − x Slope of f(t) − f(x) = lim Tangent t→x t − x

EXAMPLE 1. Find slope of the tangent line to the graph of f(x) = x2 − 4 at point (3, 5). Slope of (t2 − 4) − (x2 − 4) = lim Tangent t→x t − x t2 − x2 (t − x)(t + x) = lim = lim = 2x t→x t − x t→x t − x At x = 3, (y = 5 is irrelevant)

Slope of Tangent = 6

– Typeset by FoilTEX – 6 Velocity: How fast the position s(x) is changing.

Average h x=4 h 4h 4 = x=3 x=5 h 4 h x=2 Velocity x=0 x=1 h h 4 h 4 s(x + h) − s(x) 0 1 s = h

Instantaneous s(x + h) − s(x) = lim Velocity h→0 h

– Typeset by FoilTEX – 7 EXAMPLES

– Typeset by FoilTEX – 8 Instant. s(x + h) − s(x) = lim Velocity h→0 h

EXAMPLE 2. Motion of the particle along a line is described by s(x) = 1 + cos(x). Find instantaneous velocity at moment x = 2. Instant. s(x + h) − s(x) = lim Velocity h→0 h

EXAMPLE 2. Motion of the particle along a line is described by s(x) = 1 + cos(x). Find instantaneous velocity at moment x = 2. Instant. (1 + cos(x + h)) − (1 + cos(x)) = lim Velocity h→0 h Instant. s(x + h) − s(x) = lim Velocity h→0 h

EXAMPLE 2. Motion of the particle along a line is described by s(x) = 1 + cos(x). Find instantaneous velocity at moment x = 2. Instant. (1 + cos(x + h)) − (1 + cos(x)) = lim Velocity h→0 h cos(x + h) − cos(x) = lim h→0 h

– Typeset by FoilTEX – 9 cos(x + h) − cos(x) = lim h→0 h (using cos(x + h) = cos(x) cos(h) − sin(x) sin(h)) cos(x + h) − cos(x) = lim h→0 h (using cos(x + h) = cos(x) cos(h) − sin(x) sin(h)) cos(x) cos(h) − sin(x) sin(h) − cos(x) = lim h→0 h cos(x + h) − cos(x) = lim h→0 h (using cos(x + h) = cos(x) cos(h) − sin(x) sin(h)) cos(x) cos(h) − sin(x) sin(h) − cos(x) = lim h→0 h cos(h) − 1 sin(h) = cos(x) lim − sin(x) lim x→0 h x→0 h

= − sin(x) cos(x + h) − cos(x) = lim h→0 h (using cos(x + h) = cos(x) cos(h) − sin(x) sin(h)) cos(x) cos(h) − sin(x) sin(h) − cos(x) = lim h→0 h cos(h) − 1 sin(h) = cos(x) lim − sin(x) lim x→0 h x→0 h

= − sin(x) At x = 2, Instantaneous Velocity = − sin(2)

– Typeset by FoilTEX – 10 Compare Slope of f(t) − f(x) = lim Tangent t→x t − x Instant. s(x + h) − s(x) = lim Velocity τ→0 h Compare Slope of f(t) − f(x) = lim Tangent t→x t − x Instant. s(x + h) − s(x) = lim Velocity τ→0 h Equivalent up to the substitution

f = s, t − x = h Compare Slope of f(t) − f(x) = lim Tangent t→x t − x Instant. s(x + h) − s(x) = lim Velocity τ→0 h Equivalent up to the substitution

f = s, t − x = h

Is it the same thing?

– Typeset by FoilTEX – 11 Definition of the Derivative putting things together

– Typeset by FoilTEX – 12 Slope of Instant. Rate of Tangent Velocity Change

f(t) − f(x) s(x + h) − s(x) s(t + ∆t) − s(t) lim lim lim t→x t − x h→0 h ∆t→0 ∆t Slope of Instant. Rate of Tangent Velocity Change

f(t) − f(x) s(x + h) − s(x) s(t + ∆t) − s(t) lim lim lim t→x t − x h→0 h ∆t→0 ∆t

Definition: Derivative of f(x) at point x is f(t) − f(x) f 0(x) = lim t→x t − x  f(x + h) − f(x) or, f 0(x) = lim h→0 h

– Typeset by FoilTEX – 13 Illustration Slope of Instant. Rate of Tangent Velocity Change

y y=f(x) slope=f '(x) Rate of = Q'(t) Change x=4 h 4h 4 x=3 x=5 4 x=2 x=0 x=1 h h 4 4 c s'(x) 0 1 s Q(t) x Velocity = s'(x)

– Typeset by FoilTEX – 14 EXAMPLES

– Typeset by FoilTEX – 15 f(t) − f(x) f 0(x) = lim t→x t − x

EXAMPLE 3. Evaluate f 0(x) using definition of derivative, if f(x) = x3 + 2x. f(t) − f(x) f 0(x) = lim t→x t − x

EXAMPLE 3. Evaluate f 0(x) using definition of derivative, if f(x) = x3 + 2x. (t3 + 2t) − (x3 − 2x) f 0(x) = lim t→x t − x f(t) − f(x) f 0(x) = lim t→x t − x

EXAMPLE 3. Evaluate f 0(x) using definition of derivative, if f(x) = x3 + 2x. (t3 + 2t) − (x3 − 2x) f 0(x) = lim t→x t − x (t3 − x3) + (2t − 2x) = lim t→x t − x f(t) − f(x) f 0(x) = lim t→x t − x

EXAMPLE 3. Evaluate f 0(x) using definition of derivative, if f(x) = x3 + 2x. (t3 + 2t) − (x3 − 2x) f 0(x) = lim t→x t − x (t3 − x3) + (2t − 2x) = lim t→x t − x t3 − x3 2t − 2x = lim + lim t→x t − x t→x t − x

– Typeset by FoilTEX – 16 Consider t3 − x3 2t − 2x = lim + lim t→x t − x t→x t − x Consider t3 − x3 2t − 2x = lim + lim t→x t − x t→x t − x

(t − x)(t2 + tx + x2) 2(t − x) = lim + lim t→x (t − x) t→x (t − x)

= 1 · (x2 + x2 + x2) + 2 = 3x2 + 2

– Typeset by FoilTEX – 17 f(x + h) − f(x) f 0(x) = lim h→0 h

EXAMPLE 4. Evaluate f 0(x) using definition of √ derivative, if f(x) = x. f(x + h) − f(x) f 0(x) = lim h→0 h

EXAMPLE 4. Evaluate f 0(x) using definition of √ derivative, if f(x) = x. √ √ x + h − x f 0(x) = lim h→0 h f(x + h) − f(x) f 0(x) = lim h→0 h

EXAMPLE 4. Evaluate f 0(x) using definition of √ derivative, if f(x) = x. √ √ x + h − x f 0(x) = lim h→0 h √ √ √ √ ( x + h − x)( x + h + x) = lim √ √ h→0 h ( x + h + x) f(x + h) − f(x) f 0(x) = lim h→0 h

EXAMPLE 4. Evaluate f 0(x) using definition of √ derivative, if f(x) = x. √ √ x + h − x f 0(x) = lim h→0 h √ √ √ √ ( x + h − x)( x + h + x) = lim √ √ h→0 h ( x + h + x) (x + h) − x 1 1 = lim √ √ = √ √ = √ h→0 h( x + h + x) x + x 2 x

– Typeset by FoilTEX – 18 The Differential problem of linear approximation

– Typeset by FoilTEX – 19 Approximating function with a line

f(x+h) f(x+h) = f(x)+m·h+o(h) o(h) slope = m A

mh o(h) A where lim = 0 f(x) h h→0 h x x+h

y

Motion along a line: mh

y0 h slope=m x y = y0 + m · h 0 Approximating function with a line

f(x+h) f(x+h) = f(x)+m·h+o(h) o(h) slope = m A

mh o(h) A where lim = 0 f(x) h h→0 h x x+h

y

Motion along a line: mh

y0 h slope=m x y = y0 + m · h 0

What is m ?

– Typeset by FoilTEX – 20 Definition. Function f(x) is differentiable at x if there exists a number m such that

f(x + h) = f(x) + m · h + o(h) o(h) where lim = 0 h→0 h (o(h) “o”-little is a quantity of the order of magnitude smaller then h). Differential:

df(x)hhi = m · h

– Typeset by FoilTEX – 21 Compute m. Assume function f(x) is differenti able: f(x + h) = f(x) + m · h + o(h). Solving for m and rearranging, f(x + h) − f(x) o(h) = m + h h Compute m. Assume function f(x) is differenti able: f(x + h) = f(x) + m · h + o(h). Solving for m and rearranging, f(x + h) − f(x) o(h) = m + h h o(h) differentiability requires limh→0 h = 0 therefore, f(x + h) − f(x) lim = m h→0 h Compute m. Assume function f(x) is differenti able: f(x + h) = f(x) + m · h + o(h). Solving for m and rearranging, f(x + h) − f(x) o(h) = m + h h o(h) differentiability requires limh→0 h = 0 therefore, f(x + h) − f(x) lim = m h→0 h or f 0(x) exists and f 0(x) = m. Differentiability implies Derivative!

– Typeset by FoilTEX – 22 f(x + h) = f(x) + m · h + o(h)

EXAMPLE 5. Differentiate f(x) = sin(x). f(x + h) = f(x) + m · h + o(h)

EXAMPLE 5. Differentiate f(x) = sin(x). sin(x + h) = sin(x) + cos(x)h +[− cos(x)h + sin(x + h) − sin(x)] f(x + h) = f(x) + m · h + o(h)

EXAMPLE 5. Differentiate f(x) = sin(x). sin(x + h) = sin(x) + cos(x)h +[− cos(x)h + sin(x + h) − sin(x)]

Now, m = f 0(x) = cos(x) and o(h) = [...]. f(x + h) = f(x) + m · h + o(h)

EXAMPLE 5. Differentiate f(x) = sin(x). sin(x + h) = sin(x) + cos(x)h +[− cos(x)h + sin(x + h) − sin(x)]

Now, m = f 0(x) = cos(x) and o(h) = [...]. Notice

o(h) = cos(x)[sin(h) − h] + sin(x)[cos(h) − 1] f(x + h) = f(x) + m · h + o(h)

EXAMPLE 5. Differentiate f(x) = sin(x). sin(x + h) = sin(x) + cos(x)h +[− cos(x)h + sin(x + h) − sin(x)]

Now, m = f 0(x) = cos(x) and o(h) = [...]. Notice

o(h) = cos(x)[sin(h) − h] + sin(x)[cos(h) − 1]

Therefore, o(h) lim = 0 h→0 h

– Typeset by FoilTEX – 23 Evaluating differentials. Common notations:

df(x) = f 0(x)dx

(Compare to df(x)hhi = m · h, substituting m = f 0(x), h = dx) Evaluating differentials. Common notations:

df(x) = f 0(x)dx

(Compare to df(x)hhi = m · h, substituting m = f 0(x), h = dx) EXAMPLE 6. √ 1 d(sin x) = cos xdx, d( x) = √ dx, 2 x 1 d(arctan(x)) = dx 1 + x2

– Typeset by FoilTEX – 24