Unit 3.3 – Differentiating Inverse Functions
AP® Calculus AB & BC | The Derivative “Backwards”
Inverse function differentiation is vital for AP® Calculus: It allows you to differentiate equations in the form \(y = f^{-1}(x)\) — even when you can’t write \(y\) explicitly as a function of \(x\)! Used for logs, trig inverses, and more.
🔑 Main Formula: Inverse Functions
Key Formula (Must Memorize):
\[
(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
\]
Read as: the derivative of the inverse at \(x\) is one over the derivative of the original function, evaluated at the inverse value!
\[
\text{If } y = f^{-1}(x), \quad \frac{dy}{dx} = \frac{1}{f'(y)}
\]
Procedure for Differentiating Inverses:
- Set: \(y = f^{-1}(x)\).
- Rewrite: \(x = f(y)\).
- Differentiate both sides w.r.t. \(x\): Use implicit differentiation. (Don’t forget, \(y\) is a function of \(x\)!)
- Solve for \(\frac{dy}{dx}\).
📚 Special Cases: Derivatives of Common Inverse Functions
Inverse Trig & Log Derivatives:
- \( \frac{d}{dx} [\ln x] = \frac{1}{x}\qquad (x > 0)\)
- \( \frac{d}{dx} [\arcsin x] = \frac{1}{\sqrt{1 - x^2}} \)
- \( \frac{d}{dx} [\arccos x] = -\frac{1}{\sqrt{1 - x^2}} \)
- \( \frac{d}{dx} [\arctan x] = \frac{1}{1 + x^2} \)
- \( \frac{d}{dx} [\arccot x] = -\frac{1}{1 + x^2} \)
- \( \frac{d}{dx} [\arcsec x] = \frac{1}{|x|\sqrt{x^2 - 1}} \)
- \( \frac{d}{dx} [\arccsc x] = -\frac{1}{|x|\sqrt{x^2 - 1}} \)
All derived using the main inverse formula above!
📖 Worked Examples
Example 1: Derivative of an Inverse at a Point
If \(f(2) = 4\), and \(f'(2) = 5\), then \((f^{-1})'(4) = \frac{1}{f'(2)} = \frac{1}{5}\).
Example 2: Finding the Derivative of \(y = \arcsin(x)\)
Set \(x = \sin y\). Then \(\frac{d}{dx}[x] = \frac{d}{dx}[\sin y]\) ⇒ \(1 = \cos y\frac{dy}{dx}\).
\(\frac{dy}{dx} = \frac{1}{\cos y}\). Since \(\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}\),
So \(\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}\).
\(\frac{dy}{dx} = \frac{1}{\cos y}\). Since \(\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}\),
So \(\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}\).
Example 3: General Problem (AP®-Style Table)
Given \(f(3) = 10\), \(f'(3) = 4\), find \((f^{-1})'(10)\):
\((f^{-1})'(10) = \frac{1}{f'(3)} = \frac{1}{4}\).
\((f^{-1})'(10) = \frac{1}{f'(3)} = \frac{1}{4}\).
💡 Tips, Shortcuts, and Exam Tricks
- On AP®, look for “table of values”—find where \(f(a) = b\), then evaluate \((f^{-1})'(b) = 1 / f'(a)\).
- If given "For what value of \(x\) does \(f(x)=a\)", then \((f^{-1})'(a) = 1/f'(x)\) at that point.
- Always check: is the function invertible and \(f'(x)\neq 0\) at that point?
- For log/arithmics or trig inverses, memorize their specific derivatives.
- If you don't have an explicit inverse, just work through the table: match output-inputs carefully!
📝 Practice Problems
Try These Yourself:
- If \(f(1)=2\), \(f'(1)=7\), find \((f^{-1})'(2)\)
- Compute \(\frac{d}{dx}[\arctan x]\)
- Given \(g(3)=7\), \(g'(3)=2\), find \((g^{-1})'(7)\)
- If \(h(0)=5\), \(h'(0)=8\), what is \((h^{-1})'(5)\)?
Answers:
- \((f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{7}\)
- \(\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}\)
- \((g^{-1})'(7) = \frac{1}{g'(3)} = \frac{1}{2}\)
- \((h^{-1})'(5) = \frac{1}{8}\)
✏️ AP® Exam Success – Inverse Function Differentiation
- Box or clearly state: “By the inverse function rule: \((f^{-1})'(b) = 1 / f'(a)\) where \(f(a)=b\)”
- Show clear mapping input-to-output on tables: be methodical; AP® graders expect this step!
- Always indicate if the function is invertible and differentiable at that point.
- Log and trig inverse formulas should be instant recall: practice until they're automatic!
- Box your final answer; explain clearly if asked for justification or steps for full points.