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Inverse Functions & Derivatives

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Inverse Functions & Derivatives - Complete Interactive Lesson

Part 1: Derivative of an Inverse Function

Inverse Functions & Their Derivatives

Part 1 of 7 — The Inverse Function Derivative Formula

Topic Overview

Part	Topic
1	Inverse function derivative formula
2	Table-based inverse problems
3	Inverse trig derivatives
4	Derivatives involving $\ln$ and $\log$
5	Combining techniques
6	AP-style workshop
7	Comprehensive assessment

The Key Formula

If $g = f^{-1}$ (so $f(g(x)) = x$ ), then:

$\boxed{(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}}$

Equivalently: if $f(b) = a$ , then $(f^{-1})'(a) = \frac{1}{f'(b)}$ .

Why It Works

Differentiate $f(g(x)) = x$ by the chain rule:

$f'(g(x)) \cdot g'(x) = 1 \quad \Longrightarrow \quad g'(x) = \frac{1}{f'(g(x))}$

Key Fact: You don’t need to find $f^{-1}$ explicitly! Just find the point where $f(b)=a$ , then compute $\frac{1}{f'(b)}$ .

Step-by-Step

Step	Action
1	Find $b$ such that $f(b) = a$
2	Compute $f'(x)$

Worked Example

$f(x) = x^3 + x$ . Find $(f^{-1})'(2)$ .

Step 1: $f(b) = 2$ : $b^3+b = 2 \Rightarrow b = 1$ .

Step 2–3: $f'(x) = 3x^2+1$ . $f'(1) = 4$ .

Step 4: $(f^{-1})'(2) = \frac{1}{4}$ .

AP Tip: When $f$ is not easily invertible (like $x^3+x$ ), the formula is the only practical approach.

Practice — Inverse Derivatives 🎯

Verify understanding. 🔍

Compute. ✍️

Key Takeaways — Part 1

$(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$ — memorize this!

Part 2: Inverse Trigonometric Derivatives

Inverse Functions & Their Derivatives

Part 2 of 7 — Table-Based Inverse Problems

The AP Favorite

The AP exam frequently gives a table of $f$ and $f'$ values and asks for $(f^{-1})'$ at a point.

Part 3: eˣ and ln x Review

Inverse Functions & Their Derivatives

Part 3 of 7 — Inverse Trig Derivatives

The Formulas

$\boxed{\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}}$

Part 4: Table-Based Inverse Problems

Inverse Functions & Their Derivatives

Part 4 of 7 — Derivatives Involving $\ln$ and $\log$

Logarithmic Derivatives

$\boxed{\frac{d}{dx}[\ln x] = \frac{1}{x}} \qquad \boxed{\frac{d}{dx}[\ln u] = \frac{u'}{u}}$

Part 5: Integrals Leading to Inverse Trig

Inverse Functions & Their Derivatives

Part 5 of 7 — Combining Techniques

Mixed Problems

AP problems often combine inverse derivatives with other skills:

Combination	Example
Inverse + chain rule	$\frac{d}{dx}[\arcsin(e^x)]$

Part 6: Practice Workshop

Inverse Functions & Their Derivatives

Part 6 of 7 — AP-Style Workshop

AP FRQ Patterns Involving Inverses

Pattern	What They Ask	Key Formula
Table-based	"Find $(f^{-1})'(a)$ "

Part 7: Final Assessment

Inverse Functions & Their Derivatives

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Formula	Expression
Inverse derivative	$(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$

Find

b

where

f(b) = a

, then compute

1/f'(b)

You do NOT need to find the inverse function explicitly

Undefined when

f'(b) = 0

$\boxed{(f^{-1})'(a) = \frac{1}{f'(b)} \quad \text{where } f(b) = a}$

Strategy with Tables

Step	What to Do
1	Look up: which $b$ has $f(b) = a$ ?
2	Read $f'(b)$ from the table
3	Answer: $\frac{1}{f'(b)}$

$x$	$f(x)$	$f'(x)$
1	5	3
2	8	6
3	12	4

Find $(f^{-1})'(8)$ .

$f(2) = 8 \Rightarrow b = 2$ . $f'(2) = 6$ .

$(f^{-1})'(8) = \frac{1}{f'(2)} = \boxed{\frac{1}{6}}$

Key Fact: Students often confuse the input. $(f^{-1})'(\mathbf{8})$ asks about the $f$ -value, not the $x$ -value. Find the row where $f(x) = 8$ , NOT where $x = 8$ .

Practice — Table Problems 🎯

Identify the correct step. 🔍

Table problem. ✍️

Key Takeaways — Part 2

Table problems: look for $f(b) = a$ in the $f(x)$ column
Common mistake: using the $x$ -column instead of $f(x)$ -column
If $f'(b) = 0$ , the inverse derivative is undefined
This is one of the most common AP MC question types

$\boxed{\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}}$

$\boxed{\frac{d}{dx}[\text{arcsec}\,x] = \frac{1}{|x|\sqrt{x^2-1}}}$

Function	Derivative	Domain
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	$(-1,1)$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$
$\arctan x$	$\frac{1}{1+x^2}$	$($

Key Fact: $\arcsin$ and $\arccos$ have the same denominator but opposite signs. Memorize $\arcsin$ (positive) and $\arctan$ — the rest follow.

$\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}, \qquad \frac{d}{dx}[\arctan(u)] = \frac{u'}{1+u^2}$

$\frac{d}{dx}[\arctan(3x)]$

$= \frac{3}{1+(3x)^2} = \frac{3}{1+9x^2}$

Practice — Inverse Trig 🎯

Match the derivative. 🔍

Evaluate. ✍️

Key Takeaways — Part 3

$\frac{d}{dx}[\arcsin u] = \frac{u'}{\sqrt{1-u^2}}$ ; $\frac{d}{dx}[\arctan u] = \frac{u'}{1+u^2}$
$\arccos$ has the same formula as $\arcsin$ but negative
These formulas produce important antiderivatives: $\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$
Always apply the chain rule for compositions

$\boxed{\frac{d}{dx}[\log_a x] = \frac{1}{x\ln a}}$

Key Properties & Derivatives

Function	Derivative
$\ln x$	$1/x$
$\ln	x
$\ln(f(x))$	$f'(x)/f(x)$
$\log_a x$	$1/(x\ln a)$

Key Fact: $\frac{d}{dx}[\ln|x|] = \frac{1}{x}$ for all $x \neq 0$ . This is why $\int\frac{1}{x}\,dx = \ln|x|+C$ .

Logarithmic Differentiation

For products/quotients of many factors, take $\ln$ of both sides first:

$y = \frac{x^2\sqrt{x+1}}{(x-3)^4} \quad \Rightarrow \quad \ln y = 2\ln x + \frac{1}{2}\ln(x+1) - 4\ln(x-3)$

Differentiate: $\frac{y'}{y} = \frac{2}{x}+\frac{1}{2(x+1)}-\frac{4}{x-3}$ , then multiply by $y$ .

$\frac{d}{dx}[\ln(\sin x)]$

$= \frac{\cos x}{\sin x} = \cot x$

Practice — Logarithmic Derivatives 🎯

Evaluate. ✍️

Key Takeaways — Part 4

$\frac{d}{dx}[\ln u] = u'/u$ — the most important log derivative
$\ln$ properties simplify before differentiating: $\ln(x^n) = n\ln x$
$\int 1/x\,dx = \ln|x|+C$ (absolute value!)
Logarithmic differentiation: take $\ln$ , differentiate, multiply by $y$

Worked Example 1: Chain + Inverse Trig

$\frac{d}{dx}[\arctan(e^x)]$

$= \frac{e^x}{1+(e^x)^2} = \frac{e^x}{1+e^{2x}}$

Worked Example 2: Implicit + ln

$\ln y + xy = 5$ . Find $\frac{dy}{dx}$ .

$\frac{1}{y}\cdot y' + y + xy' = 0$ $y'\left(\frac{1}{y}+x\right) = -y$ $y' = \frac{-y}{\frac{1}{y}+x} = \frac{-y^2}{1+xy}$

Worked Example 3: Related Antiderivatives

$\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C \qquad \int\frac{dx}{1+x^2} = \arctan x + C$

AP Tip: These antiderivatives often appear in definite integral MC problems. Evaluate $\arcsin$ or $\arctan$ at the limits.

Practice — Mixed Problems 🎯

Classify the technique. 🔍

Evaluate. ✍️

Key Takeaways — Part 5

Chain rule + inverse trig: outer is $\arcsin/\arctan$ , inner is $u$
$\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$ , $\int\frac{dx}{1+x^2} = \arctan x + C$
$\ln$ + implicit: differentiate both sides, solve for $y'$
AP loves definite integrals using $\arcsin$ and $\arctan$

\frac{1}{f ^{'} ( f ^{- 1} ( a ))}

Full Worked AP Problem

Let $f$ be a differentiable function with $f(3) = 7$ , $f'(3) = 4$ , $f(7) = 1$ , and $f'(7) = -2$ . Let $g = f^{-1}$ .

(a) Find $g'(7)$ .

(b) Find the equation of the tangent line to $y = g(x)$ at $x = 7$ .

(c) Find $h'(7)$ where $h(x) = \ln(g(x))$ .

Solution (a): $g'(7) = \frac{1}{f'(g(7))}$ . Since $f(3) = 7$ , $g(7) = 3$ . Thus $g'(7) = \frac{1}{f'(3)} = \frac{1}{4}$ .

Solution (b): Point: $(7, g(7)) = (7, 3)$ . Slope: $g'(7) = 1/4$ . $y - 3 = \frac{1}{4}(x - 7) \quad \Rightarrow \quad y = \frac{1}{4}x + \frac{5}{4}$

Solution (c): $h'(x) = \frac{g'(x)}{g(x)}$ . Thus $h'(7) = \frac{1/4}{3} = \frac{1}{12}$ .

AP Tip: Part (c) combines the chain rule with $\ln$ applied to an inverse function — a common multi-step problem.

AP-Style Practice 🎯

Identify the correct value. 🔍

Multi-step AP problem. ✍️

Key Takeaways — Part 6

AP FRQs: always start by finding $f^{-1}(a)$ from the table/given info
Tangent line to inverse: use the point $(a, f^{-1}(a))$ and slope $1/f'(f^{-1}(a))$
Combining $\ln$ with inverse: apply chain rule carefully
Justifications: show $f' > 0$ or $f' < 0$ everywhere to prove invertibility

Mistake	Correct Approach
Using $f'(a)$ instead of $f'(f^{-1}(a))$	Always find $f^{-1}(a)$ first
Forgetting chain rule on $\arcsin(u)$	Multiply by $u'$
Sign error on $\arccos$	$\arccos$ has a negative: $-u'/\sqrt{1-u^2}$
$\ln(fg) \neq \ln f \cdot \ln g$	Use $\ln(fg) = \ln f + \ln g$
Swapping point and slope	Inverse swaps $x$ / $y$ : point is $(a, f^{-1}(a))$

Quiz Set 1 — Core Concepts 🎯

Quiz Set 2 — Advanced Applications 🎯

Match correctly. 🔍

Final Challenge. ✍️

🎉 Topic Complete!

You've mastered Inverse Functions & Their Derivatives:

Part	Topic	Status
1	Inverse derivative formula	✅
2	Table-based problems	✅
3	Inverse trig derivatives	✅
4	Logarithmic derivatives	✅
5	Combining techniques	✅
6	AP-style workshop	✅
7	Comprehensive assessment	✅

Key Fact: The formula $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$ is the single most common inverse derivative question on the AP exam. Master the three-step process: (1) find $f^{-1}(a)$ , (2) evaluate $f'$ there, (3) take the reciprocal.

(- \infty, \infty)

Inverse Functions & Derivatives

Strategy with Tables

Worked Example

Common Pitfall

Key Takeaways — Part 2

Complete Reference

With the Chain Rule

Worked Example

Key Takeaways — Part 3

Key Properties & Derivatives

Logarithmic Differentiation

Worked Example

Key Takeaways — Part 4

Worked Example 1: Chain + Inverse Trig

Worked Example 2: Implicit + ln

Worked Example 3: Related Antiderivatives

Key Takeaways — Part 5

Full Worked AP Problem

Key Takeaways — Part 6

Common AP Mistakes

🎉 Topic Complete!