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