Derivatives of Inverse Functions
If $y = f^{-1}(x)$, flipping the fraction gives $\dfrac{dy}{dx} = \dfrac{1}{\,f'(y)\,}$. This elegant rule unlocks the derivatives of $\sin^{-1}x$, $\cos^{-1}x$ and $\tan^{-1}x$ — three functions that appear constantly in HSC integration. In this lesson you'll derive the general rule, practise it step-by-step, and see exactly how it connects to what you already know about the chain rule.
You know $\dfrac{d}{dx}(\sin x) = \cos x$. Without looking ahead — write what you think $\dfrac{d}{dx}(\sin^{-1} x)$ might be and explain your reasoning.
When we invert a function, the graph reflects across the line $y = x$. This swaps the roles of $x$ and $y$ — and so it swaps $\dfrac{dx}{dy}$ and $\dfrac{dy}{dx}$. The key formula is:
If $x = f(y)$, then $\dfrac{dx}{dy} = f'(y)$. But we want $\dfrac{dy}{dx}$ — and by the inverse function theorem:
$$\frac{dy}{dx} = \frac{1}{\,dx/dy\,} = \frac{1}{f'(y)}$$
Equivalently, if $y = f^{-1}(x)$, then $\dfrac{dy}{dx} = \dfrac{1}{f'(f^{-1}(x))}$.
Key facts
- If $y = f^{-1}(x)$ then $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$
- Equivalently $\dfrac{dy}{dx} \cdot \dfrac{dx}{dy} = 1$
- The derivative fails where $f'(y) = 0$
Concepts
- Why reflecting across $y = x$ swaps the derivative and its reciprocal
- How to convert the result into a function of $x$ only
- The connection to implicit differentiation and the chain rule
Skills
- Apply $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$ to find derivatives of inverses
- Express derivatives in terms of $x$ using the inverse substitution
- Identify where the inverse derivative is undefined
Setting up: Suppose $y = f^{-1}(x)$. Rewrite this as $x = f(y)$. Now differentiate both sides with respect to $x$ using the chain rule:
This is valid whenever $f'(y) \neq 0$. To express the result purely in terms of $x$, substitute $y = f^{-1}(x)$:
Preview: For $f(y) = \sin y$ on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$: $f'(y) = \cos y$, so
$\dfrac{d}{dx}(\sin^{-1} x) = \dfrac{1}{\cos(\sin^{-1} x)} = \dfrac{1}{\sqrt{1-x^2}}$
using the Pythagorean identity $\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}$. We derive this fully in Lesson 10.
Setting up: Suppose $y = f^{-1}(x)$. Rewrite this as $x = f(y)$. Now differentiate both sides with respect to $x$ using the chain rule:
Pause — copy the inverse-function derivative rule: $y=f^{-1}(x)\Rightarrow\frac{dy}{dx}=\frac{1}{f'(y)}$ with the derivation via $x=f(y)$ and implicit differentiation into your book.
Quick check: If $y = f^{-1}(x)$ and $f'(y) = 3$, what is $\dfrac{dy}{dx}$?
We just saw the inverse-function derivative rule: if $y=f^{-1}(x)$ then rewrite as $x=f(y)$, differentiate implicitly to get $1=f'(y)\frac{dy}{dx}$, so $\frac{dy}{dx}=\frac{1}{f'(y)}$. That raises a question: the answer $\frac{1}{f'(y)}$ still involves $y$ — how do you express it in terms of $x$ when $y=f^{-1}(x)$? This card answers it → substitute back: replace every $y$ with $f^{-1}(x)$, or use a right-triangle diagram if $f$ is trigonometric.
After finding $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$, the answer still involves $y$. The final step is to substitute $y = f^{-1}(x)$ and simplify, often using a Pythagorean identity.
General method:
- Write $x = f(y)$ and differentiate: $1 = f'(y)\,\dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$.
- Draw a right-triangle labelled with sides reflecting $y = f^{-1}(x)$ to identify the remaining side.
- Express $f'(y)$ in terms of $x$ using the triangle or Pythagorean identity.
- State the final derivative as a function of $x$ only.
After finding $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$, the answer still involves $y$. The final step is to substitute $y = f^{-1}(x)$ and simplify, often using a Pythagorean identity.
Pause — copy the back-substitution step: replace $y$ with $f^{-1}(x)$ in $\frac{1}{f'(y)}$ to express the derivative entirely in terms of $x$ into your book.
Did you get this? True or false: to differentiate $y = f^{-1}(x)$, you rewrite as $x = f(y)$ and differentiate both sides with respect to $x$.
Worked examples · 3 in a row, reveal as you go
If $f(x) = x^3 + x$, find $\dfrac{d}{dx}[f^{-1}(x)]$ at $x = 2$.
Use the inverse function rule to find $\dfrac{d}{dx}(\ln x)$, given that the inverse of $\ln x$ is $e^x$.
Use the inverse function rule to find $\dfrac{d}{dx}(\cos^{-1} x)$.
Fill the gap: If $x = e^y$, then $\dfrac{dx}{dy} =$ , so $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the derivative of $\cos^{-1}x$ is $\dfrac{1}{\sqrt{1-x^2}}$ (positive).
Activities · practice with the ideas
Let $f(x) = x^5 + 2x$. Find $[f^{-1}]'(3)$. (Hint: find $y$ such that $y^5 + 2y = 3$ first.)
Use $x = \tan y$ (with $y \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$) to derive $\dfrac{d}{dx}(\tan^{-1}x)$.
Explain why $\dfrac{d}{dx}(\sin^{-1}x)$ is undefined at $x = \pm 1$.
Verify: $\dfrac{d}{dx}(\sin^{-1}x) + \dfrac{d}{dx}(\cos^{-1}x) = 0$. What does this say about $\sin^{-1}x + \cos^{-1}x$?
If $g$ is differentiable and $g(g^{-1}(x)) = x$ for all $x$, use the chain rule on both sides to derive $[g^{-1}]'(x) = \dfrac{1}{g'(g^{-1}(x))}$.
Odd one out: Three of these derivatives are correct. Which one is NOT?
Earlier you estimated what $\dfrac{d}{dx}(\sin^{-1}x)$ might be.
The actual answer is $\dfrac{1}{\sqrt{1-x^2}}$. The key insight: rewriting $y = \sin^{-1}x$ as $x = \sin y$ and differentiating implicitly gives $1 = \cos y \cdot \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{1}{\cos y} = \dfrac{1}{\sqrt{1-x^2}}$. Were you on the right track?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the inverse function derivative rule: if $y = f^{-1}(x)$, then $\dfrac{dy}{dx} = $ … (1 mark)
Q2. Use the inverse function rule to find $[f^{-1}]'(8)$ given $f(x) = x^3$. (2 marks)
Q3. Starting from $x = \tan y$, derive $\dfrac{d}{dx}(\tan^{-1}x) = \dfrac{1}{1+x^2}$. Show all working. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $f'(y) = 5y^4+2$. At $y=1$: $f(1)=3$ ✓. $[f^{-1}]'(3) = \dfrac{1}{5+2} = \dfrac{1}{7}$.
2. $x = \tan y \Rightarrow \dfrac{dx}{dy}=\sec^2 y \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sec^2 y}=\cos^2 y$. Since $\tan y = x$, hypotenuse $=\sqrt{1+x^2}$, so $\cos y = \dfrac{1}{\sqrt{1+x^2}}$ and $\cos^2 y = \dfrac{1}{1+x^2}$. Hence $\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^2}$.
3. At $x=\pm1$, $\cos y = \sqrt{1-\sin^2 y}=\sqrt{1-x^2}=0$, so the denominator is zero and the derivative is undefined (vertical tangent on the original $\sin$ curve).
4. Sum of derivatives $= \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-x^2}} = 0$, confirming $\sin^{-1}x + \cos^{-1}x = C$ (a constant). That constant is $\dfrac{\pi}{2}$.
5. Differentiate $g(g^{-1}(x))=x$: $g'(g^{-1}(x)) \cdot [g^{-1}]'(x) = 1$, so $[g^{-1}]'(x) = \dfrac{1}{g'(g^{-1}(x))}$.
Q1 (1 mark): $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$ [1].
Q2 (2 marks): $f(y)=8 \Rightarrow y=2$ [1]. $f'(y)=3y^2$, at $y=2$: $f'(2)=12$. $[f^{-1}]'(8)=\dfrac{1}{12}$ [1].
Q3 (3 marks): $x=\tan y \Rightarrow \dfrac{dx}{dy}=\sec^2 y$ [1]. $\dfrac{dy}{dx}=\dfrac{1}{\sec^2 y}$ [1]. Use $\sec^2 y = 1+\tan^2 y = 1+x^2$, so $\dfrac{dy}{dx}=\dfrac{1}{1+x^2}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inverse function questions. Lighter alternative to the boss.
Mark lesson as complete
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