Derivatives of Exponential and Logarithm Functions
10/17/2011 e0+h − e0 eh − 1 lim = lim = 1 h→0 h h→0 h
The Derivative of y = ex
Recall! ex is the unique exponential function whose slope at x = 0 is 1:
m=1 The Derivative of y = ex
Recall! ex is the unique exponential function whose slope at x = 0 is 1:
m=1
e0+h − e0 eh − 1 lim = lim = 1 h→0 h h→0 h ex eh − 1 = lim h→0 h eh − 1 = ex lim h→0 h = ex ∗ 1
So d ex = ex dx
d ex+h − ex ex = lim dx h→0 h
The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h ex eh − 1 = lim h→0 h eh − 1 = ex lim h→0 h = ex ∗ 1
So d ex = ex dx
The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h
d ex+h − ex ex = lim dx h→0 h eh − 1 = ex lim h→0 h = ex ∗ 1
So d ex = ex dx
The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h
d ex+h − ex ex = lim dx h→0 h ex eh − 1 = lim h→0 h = ex ∗ 1
So d ex = ex dx
The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h
d ex+h − ex ex = lim dx h→0 h ex eh − 1 = lim h→0 h eh − 1 = ex lim h→0 h So d ex = ex dx
The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h
d ex+h − ex ex = lim dx h→0 h ex eh − 1 = lim h→0 h eh − 1 = ex lim h→0 h = ex ∗ 1 The Derivative of y = ex ...
eh − 1 lim = 1 h→0 h
d ex+h − ex ex = lim dx h→0 h ex eh − 1 = lim h→0 h eh − 1 = ex lim h→0 h = ex ∗ 1
So d ex = ex dx The Chain Rule
Theorem Let u be a function of x. Then d du eu = eu . dx dx = 17e17x
= cos(x)esin x
√ 2 = √2x+1 e x +1 2 x2+x
Notice, every time: d ef (x) = f 0(x)ef (x) dx
Examples
Calculate... d 17x 1. dx e
d sin x 2. dx e √ d x2+x 3. dx e Notice, every time: d ef (x) = f 0(x)ef (x) dx
Examples
Calculate... d 17x 17x 1. dx e = 17e
d sin x sin x 2. dx e = cos(x)e √ √ 2 2 3. d e x +x = √2x+1 e x +1 dx 2 x2+x Examples
Calculate... d 17x 17x 1. dx e = 17e
d sin x sin x 2. dx e = cos(x)e √ √ 2 2 3. d e x +x = √2x+1 e x +1 dx 2 x2+x
Notice, every time: d ef (x) = f 0(x)ef (x) dx 1 1 = = eln(x) x
Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 = dx ey
d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! 1 1 = = eln(x) x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 = dx ey
d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x 1 1 = = eln(x) x
So dy 1 = dx ey
d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx 1 1 = = eln(x) x
d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 = dx ey 1 = x
d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 1 = = dx ey eln(x) d 1 dx ln(x) = x
The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 1 1 = = = dx ey eln(x) x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =⇒ ey = x
Take a derivative of both sides of ey = x to get dy ey = 1 dx
So dy 1 1 1 = = = dx ey eln(x) x
d 1 dx ln(x) = x Does it make sense?
d 1 dx ln(x) = x
1
f (x) = ln(x) 1 2 3 4
-1
3
2 1 f (x) = x
1
1 2 3 4 2x 2 = = x2 x
2x cos(x2) = sin(x2)
Notice, every time:
d f 0(x) ln(f (x)) = dx f (x)
Examples
Calculate d 2 1. dx ln x
d 2 2. dx ln(sin(x )) Notice, every time:
d f 0(x) ln(f (x)) = dx f (x)
Examples
Calculate 2x 2 1. d ln x2 = = dx x2 x
2x cos(x2) 2. d ln(sin(x2)) = dx sin(x2) Examples
Calculate 2x 2 1. d ln x2 = = dx x2 x
2x cos(x2) 2. d ln(sin(x2)) = dx sin(x2)
Notice, every time:
d f 0(x) ln(f (x)) = dx f (x) d = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx
(ln(2) is a constant!!!)
For example: d 2x dx
d 1 You try: log (x) = dx 2 ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a d = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx
(ln(2) is a constant!!!)
d 1 You try: log (x) = dx 2 ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d 2x dx = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x
(ln(2) is a constant!!!)
d 1 You try: log (x) = dx 2 ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d d 2x = ex ln(2) dx dx = ln(2) ∗ 2x
d 1 You try: log (x) = dx 2 ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) dx dx
(ln(2) is a constant!!!) d 1 You try: log (x) = dx 2 ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx
(ln(2) is a constant!!!) 1 = ln(2) ∗ x
The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx
(ln(2) is a constant!!!)
d You try: log (x) dx 2 The Calculus Standards: ex and ln x
To get the other derivatives:
ax = ex ln a ln x log x = a ln a
For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx
(ln(2) is a constant!!!)
d 1 You try: log (x) = dx 2 ln(2) ∗ x 1. If k = 1, then y = ex has this property and thus solves the equation. 2. For any k, y = ekx solves the equation too!
d This equation, dx y = ky is an example of a differential equation.
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y?? 2. For any k, y = ekx solves the equation too!
d This equation, dx y = ky is an example of a differential equation.
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y?? 1. If k = 1, then y = ex has this property and thus solves the equation. d This equation, dx y = ky is an example of a differential equation.
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y?? 1. If k = 1, then y = ex has this property and thus solves the equation. 2. For any k, y = ekx solves the equation too! Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y?? 1. If k = 1, then y = ex has this property and thus solves the equation. 2. For any k, y = ekx solves the equation too!
d This equation, dx y = ky is an example of a differential equation.