AP Calculus - 7.1 Inverse Functions Name:_______________
Part I: Find the derivative of a function’s inverse at a given point.
Theorem: If f is a one-to-one differentiable function with inverse function 𝑓−1 and 𝑓′(𝑓−1(𝑦)) ≠ 0, then the inverse
function is differentiable at 𝑦 and…
(𝒇−𝟏)′(𝒚) =𝟏
𝒇′(𝒇−𝟏(𝒚)) Note: 𝑓(𝑥) = 𝑦 and 𝑓−1(𝑦) = 𝑥
2 options:
1) Find the inverse and then calculate the derivative at the given y-value (the y-value is the given number)
or
2) Use the above formula
a. Differentiate 𝑓(𝑥) (not the inverse) and evaluate at the x-value (not the given number)
b. Take the reciprocal
Example: If 𝑓(𝑥) = 𝑥2 + 1, 𝑥 ≥ 0 find (𝑓−1)′(5).
Complete the following 2 practice problems.
Problem 1: If 𝑓(𝑥) = 𝑥3 find (𝑓−1)′(27).
Problem 2: If 𝑔(𝑥) = √𝑥 − 2, find (𝑔−1)′(3).
Note
Problem 3: If 𝑓(𝑥) = 𝑥3 find (𝑓−1)′(8).
Problem 4: If 𝑔(𝑥) = √𝑥 − 2, find (𝑔−1)′(2).
Problem 5: 𝑓(𝑥) = 9 − 𝑥2, 0 ≤ 𝑥 ≤ 3 find (𝑓−1)′(8).
Problem 6: If 𝑔(𝑥) = sin 𝑥, −
𝜋
2≤ 𝑥 ≤
𝜋
2,
find (𝑔−1)′ (√2
2).
Problem 7: If 𝑓(𝑥) = 2𝑥3 + 3𝑥 − 4 find (𝑓−1)′(−4).
Problem 8: Suppose 𝑓−1 is the inverse function of f and
𝑓(2) = 7, 𝑓′(2) =3
5. Find (𝑓−1)′(7). Write the tangent line
equation to 𝑓−1 at the point 𝑓(2) = 7
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Things to Remember from Pre-Calculus:
A relation is a function if the graph passes the “Vertical Line Test”
The “Horizontal Line Test” determines if a function will be one-to-one.
A function is “one-to-one” if and only if the graph passes both the HLT and VLT
A function must be one-to-one in order to have an inverse.
Calculating an inverse:
1. Switch 𝒙 & 𝒚
2. Solve for 𝒚
Inverses are symmetrical about the line 𝒚 = 𝒙
𝒇−𝟏(𝒙) 𝐝𝐨𝐞𝐬 𝒏𝒐𝒕 𝐦𝐞𝐚𝐧𝟏
𝒇(𝒙)
AP Calculus - 7.2 Exponential Functions and Their Derivatives Name: __________________
Definition of 𝒆 (Euler’s constant after Leonhard Euler 1707-1783)
* 𝑒 is a constant and is approximately 2.71828182845904523536 …. *
𝑒 can also be found using 𝐥𝐢𝐦𝒏→∞
(𝟏 +𝟏
𝒏)
𝒏
e is also used as the base in the natural logarithm and the compounding continuously interest formula 𝑷𝒆𝒓𝒕.
Derivative of 𝒆𝒙
𝑑
𝑑𝑥(𝑒𝑥) = 𝑒𝑥
𝑑
𝑑𝑥(𝑒𝑢) = 𝑒𝑢
𝑑𝑢
𝑑𝑥
This means that the function 𝑓(𝑥) = 𝑒𝑥 has the property that it is its own derivative! The geometrical significance of this
fact is that the slope of a tangent line to the curve 𝑦 = 𝑒𝑥 at any point is equal to the y-coordinate of the point.
Example 1: Differentiate the functions
Function Derivative
𝑦 = 𝑒5𝑥
𝑦 = 𝑒tan 𝑥
𝑦 = 𝑒−4𝑥 sin 5𝑥
𝑦 = 𝑥𝑒−𝑥
Integral of 𝒆𝒙
Example 2: Evaluate each integral.
Function Integral
∫ 14𝑒7𝑥 𝑑𝑥
∫ 𝑥2𝑒𝑥3𝑑𝑥
∫ 𝑒−3𝑥𝑑𝑥
1
0
∫ 𝑠𝑖𝑛 𝑥 𝑒𝑐𝑜𝑠 𝑥 𝑑𝑥
Example 3: Let R be the region between the graphs of 𝑦 = 𝑒2𝑥, 𝑦 = 1 and 𝑥 = 3.
(a) Calculate the area of R.
(b) Calculate the volume of the solid when R is revolved about the x-axis.
(c) Set-up, but do not evaluate an integral expression for the volume of the solid when R is revolved about the y-axis.
AP Calculus - 7.3 Logarithmic Equations Name: ___________________
Properties of Logs
Rewriting exponential/logarithmic equations
Solving exponential equations
Solving logarithmic equations
AP Calculus - 7.4 Derivatives and Integrals of Exponential Functions Name:______________
*Chain rule always applies*
Examples:
Calculate the Derivatives
Evaluate the integrals
𝑑
𝑑𝑥(𝑎𝑥) = 𝑎𝑥 ln 𝑎
𝑑
𝑑𝑥(𝑒𝑥) = 𝑒𝑥
∫ 𝑎𝑥𝑑𝑥 =𝑎𝑥
ln 𝑎 + 𝐶 𝑎 ≠ 1
𝑑
𝑑𝑥(7𝑥)
𝑑
𝑑𝑥(33𝑥 )
𝑑
𝑑𝑥(5cos 𝑥)
𝑑
𝑑𝑥(5𝑒2𝑥
)
𝑑
𝑑𝑥(8𝑠𝑖𝑛𝑥)
𝑑
𝑑𝑥(6𝑒tan(𝑥 ))
𝑑
𝑑𝑥(
3𝑥
𝑒𝑥 )
∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶
AP Calculus - 7.4 Derivatives and Integrals of Logarithmic Functions Name:______________
𝑑
𝑑𝑥 (ln 𝑥) =
1
𝑥
𝑑
𝑑𝑥 (ln 𝑢) =
1
𝑢
𝑑𝑢
𝑑𝑥
Proof:
Examples: Calculate 𝑑
𝑑𝑥 Examples: Evaluate the following integrals
ln(sin 𝑥)
ln √𝑥 + 1
𝑥2 ln 𝑥
ln(𝑥2 + 10)
sin(ln 𝑥)
𝐥𝐧 |𝒙|
𝒅
𝒅𝒙 (𝐥𝐧|𝒙|) =
𝟏
𝒙
∫ 𝐭𝐚𝐧 𝒙 𝒅𝒙 = 𝐥𝐧 | 𝐬𝐞𝐜 𝒙| + 𝑪
Proof:
∫1
𝑥2 𝑑𝑥
2
1
∫1
𝑥 𝑑𝑥
2
1
∫𝑥
𝑥2 + 1 𝑑𝑥
∫ln 𝑥
𝑥 𝑑𝑥
𝑒
1
∫ 𝐭𝐚𝐧 𝒙 𝒅𝒙
Derivatives of general logarithms
Examples: Find 𝑑
𝑑𝑥
log10(2 + sin 𝑥)
2𝑥 log4 √𝑥
log2(cos(𝜋𝑥))
𝑑
𝑑𝑥(log𝑎 𝑥) =
1
𝑥 ln 𝑎
AP Calculus - 7.4 Logarithmic Differentiation Name:______________
Logarithmic Differentiation:
Can be used for complicated functions, but mostly used when you have a variable as a base and an exponent.
1) Take natural logs of both sides
2) Differentiate implicitly
3) Solve for 𝑦′ or 𝑑𝑦
𝑑𝑥
Find 𝑑𝑦
𝑑𝑥 𝑦 = 𝑥√𝑥
Find 𝑑𝑦
𝑑𝑥 𝑦 =
𝑥34 √𝑥2+ 1
(3𝑥+2)5
AP Calculus - 7.6 Derivatives of Inverse Trig Functions Name:______________
sin−1 𝑥 = arcsin 𝑥 𝑐𝑜𝑠−1 𝑥 = arccos 𝑥 tan−1 𝑥 = arctan 𝑥
Purpose: Used to calculate unknown angles
Example 1: sin 𝑥 =√3
2 => 𝑥 = sin−1 √3
2 => 𝑥 = 60° =
𝜋
3 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Equation Solution
𝒄𝒐𝒔−𝟏 (𝟏/𝟐) = 𝒙
𝒂𝒓𝒄𝒕𝒂𝒏 √𝟑 = 𝟒𝒙
𝒙 = 𝒔𝒊𝒏−𝟏(𝒔𝒊𝒏𝟕𝝅
𝟔)
How to calculate the derivative of 𝒚 = 𝐬𝐢𝐧−𝟏 𝒙.
If 𝑦 = sin−1 𝑥, then sin 𝑦 = 𝑥. Draw a right triangle. Use the Pythagorean Theorem to solve for the adjacent side,
The rest of the derivatives of the inverse trigonometric functions can be done the same way.
𝑦
1
𝑥
√1 − 𝑥2
𝑑
𝑑𝑥(𝑥 = sin 𝑦)
1 = cos 𝑦 ∙𝑑𝑦
𝑑𝑥
1
cos 𝑦 =
𝑑𝑦
𝑑𝑥
𝒅𝒚
𝒅𝒙=
𝟏
√𝟏 − 𝒙𝟐
d
dx(𝐬𝐢𝐧−𝟏 𝒙) =
1
√1−𝑥2
d
dx(𝐜𝐬𝐜−𝟏 𝒙) =
−1
𝑥 √𝑥2− 1
d
dx(𝐜𝐨𝐬−𝟏 𝒙) =
−1
√1−𝑥2
d
dx(𝐬𝐞𝐜−𝟏 𝒙) =
1
𝑥 √𝑥2− 1
d
dx(𝐭𝐚𝐧−𝟏 𝒙) =
1
1+𝑥2
d
dx(𝐜𝐨𝐭−𝟏 𝒙) =
−1
1+𝑥2
Function Derivative
𝒇(𝒙) = 𝒄𝒐𝒔−𝟏 (𝟐𝒙)
𝒇(𝒙) = 𝒍𝒏 (𝒂𝒓𝒄𝒕𝒂𝒏 𝒙)
𝒇(𝒙) = 𝒆𝒂𝒓𝒄𝒔𝒊𝒏𝒙
Integral Integral Evaluated
∫𝟓
√𝟏 − 𝒙𝟐 𝒅𝒙
∫−𝟑
√𝟏 − 𝒙𝟐
√𝟐/𝟐
𝟎
𝒅𝒙
∫𝟏
𝟏 + (𝒙𝟐)
𝟐
𝟐
𝟎
𝒅𝒙
∫𝟏
𝟗 + 𝒙𝟐
√𝟑
𝟎
𝒅𝒙
AP Calculus - 7.8 Indeterminant Forms and l’Hospital’s Rule Name:______________
The following are called indeterminant forms of limits:
Division Multiplication Subtraction Power
0
0
0 ∙ ∞
∞ − ∞
00
1∞
∞
∞
Can be changed to division
Must change to a fraction
∞0
When evaluating a limit, if you get an indeterminant form you have a couple of options:
1) Try to factor and cancel the “problem factor”
2) Use l’Hospital’s rule, but you can only use this rule if you encounter an indeterminant form!
l’Hospital’s Rule (Marquiz de l’Hospital – 1661-1704) (pronounced “la-hope-e-tall”)
Suppose f and g are differentiable and 𝑔′(𝑥) ≠ 0 on an open interval I that contains a (except possible at𝑎). Suppose that
the lim𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥) yields an indeterminant form, then
𝐥𝐢𝐦𝒙→𝒂
𝒇(𝒙)
𝒈(𝒙)= 𝐥𝐢𝐦
𝒙→𝒂
𝒇′(𝒙)
𝒈′(𝒙)
Evaluate the limit
𝐥𝐢𝐦𝒙→𝟎
𝟐𝒙
𝒙
𝐥𝐢𝐦𝒙→𝟏
𝒙𝟐 − 𝟐𝒙 + 𝟏
𝒙 − 𝟏
𝐥𝐢𝐦𝒙→−𝟐
𝒙𝟑 + 𝟖
𝒙 + 𝟐
𝐥𝐢𝐦𝒙→𝟏
𝒍𝒏 𝒙
𝒙 − 𝟏
𝐥𝐢𝐦𝒙→∞
𝒆𝒙
𝒙𝟐
𝐥𝐢𝐦𝒙→∞
𝒍𝒏 𝒙
√𝒙𝟑
𝐥𝐢𝐦𝒙→𝝅−
𝒔𝒊𝒏 𝒙
𝟏 − 𝒄𝒐𝒔 𝒙
𝐥𝐢𝐦𝒙→𝟎
𝒙𝒙
𝐥𝐢𝐦𝒙→∞
𝒆𝒙
𝒙𝟑