Inverse Functions & Derivatives

Inverse Functions & Their Derivatives

Part 1 of 7 — The Inverse Function Derivative Formula

Topic Overview

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

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!
• Find $b$ where $f(b) = a$, then compute $1/f'(b)$
• You do NOT need to find the inverse function explicitly
• Undefined when $f'(b) = 0$

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.

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

Strategy with Tables

Worked Example

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

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

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

Common Pitfall

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

Inverse Functions & Their Derivatives

Part 3 of 7 — Inverse Trig Derivatives

The Formulas

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

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

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

Complete Reference

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

With the Chain Rule

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

Worked Example

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

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

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

Inverse Functions & Their Derivatives

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

Logarithmic Derivatives

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

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

Key Properties & Derivatives

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 $y$.

Worked Example

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

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

Practice — Logarithmic Derivatives 🎯

Identify. 🔍

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 $y$

Inverse Functions & Their Derivatives

Part 5 of 7 — Combining Techniques

Mixed Problems

AP problems often combine inverse derivatives with other skills:

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 $\arctan$

Inverse Functions & Their Derivatives

Part 6 of 7 — AP-Style Workshop

AP FRQ Patterns Involving Inverses

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

Inverse Functions & Their Derivatives

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Common AP Mistakes

Quiz Set 1 — Core Concepts 🎯

Quiz Set 2 — Advanced Applications 🎯

Match correctly. 🔍

Final Challenge. ✍️

🎉 Topic Complete!

You've mastered Inverse Functions & Their Derivatives:

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.