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Derivatives of Inverse Functions

Every invertible function has an inverse that undoes it — and you can get the inverse’s derivative almost for free, by reflection. The headline payoff is the inverse-trig functions \(\arcsin,\arccos,\arctan\), whose derivatives come out purely algebraic. This lesson leans on both the chain rule and implicit differentiation.

The graph of \(f^{-1}\) is the graph of \(f\) reflected across the line \(y=x\). That reflection swaps \(x\) and \(y\) — and it turns a tangent of slope \(m\) into one of slope \(1/m\). So the inverse’s derivative is the reciprocal of the original’s, measured at the mirror point: \(\big(f^{-1}\big)'(x)=\dfrac{1}{f'\!\big(f^{-1}(x)\big)}\). Apply that to \(\sin,\cos,\tan\) and the inverse-trig derivatives fall out — with no trig left in the answer.

An inverse is a reflection

Here is \(f(x)=e^{x}\) in orange and its inverse \(f^{-1}(x)=\ln x\) in blue — mirror images across the dashed line \(y=x\). Drag the point. Wherever \(f\) has slope \(m\), its mirror on \(f^{-1}\) has slope \(1/m\): the reflection literally swaps rise and run.

drag the input \(a\)

slope of \(f\) at \(a\)

slope of \(f^{-1}\) at the mirror point

their product

Why the slope flips

Reflecting a line across \(y=x\) swaps its rise and run, so a slope of \(m\) becomes \(1/m\). That single fact is the whole rule. There’s a one-line proof too: if \(y=f^{-1}(x)\) then \(x=f(y)\); differentiate both sides — that’s implicit differentiation — to get the same thing.

One caveat the picture makes obvious: where \(f'(a)=0\) the mirror tangent is vertical, so \(f^{-1}\) isn’t differentiable there.

A worked inverse-value problem

This is the exact shape of the most common AP question on inverses. Suppose \(f\) is invertible with \(f(2)=5\) and \(f'(2)=3\). Find \(\big(f^{-1}\big)'(5)\). Press step.

press step

The inverse-trig derivatives

Where does \(\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}\) actually come from? Implicit differentiation — the same move from the last lesson. Watch it derive itself.

The three you need, side by side. Notice \(\arccos\) is just the negative of \(\arcsin\), and none of them leave any trig behind:

→

The slope of \(\arctan\) is pure algebra

\(\arctan x\) climbs from \(-\tfrac{\pi}{2}\) to \(\tfrac{\pi}{2}\) and levels off. Its slope is \(\dfrac{1}{1+x^2}\) — the clean blue bump below, peaking at \(1\) and never negative. Drag \(x\) and read the slope straight off the curve: no trig anywhere.

point \(x\)

at \(x\)

height \(\arctan x\)

slope \(\dfrac{1}{1+x^2}\)

Differentiate the inverse-trig functions

Each one is an inverse-trig outer wrapped around an inner — so the chain rule rides along. Pick the derivative.

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Differentiate with respect to \(x\):

The whole idea on one card

Reflect \(f\) across \(y=x\) to get \(f^{-1}\) — the mirror image

The reflection swaps rise and run, so slopes reciprocate

Evaluate \(f'\) at the matching input \(f^{-1}(x)\), then take the reciprocal

For inverse-trig, the same rule gives clean algebraic derivatives

The ideas everything else builds on

Reflect across \(y=x\)

The inverse is the mirror image of \(f\); inputs and outputs trade places.

Slopes reciprocate

A tangent of slope \(m\) reflects to one of slope \(1/m\).

Match the point first

Use \(f'\) at \(f^{-1}(x)\), not at \(x\) itself.

Inverse-trig is algebraic

\(\arcsin,\arccos,\arctan\) all differentiate to expressions with no trig.

Your turn

Quick quiz

Eight fast checks across the whole lesson.

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Common mistakes & exam tips

Dropping the chain rule on inverse-trig

\(\frac{d}{dx}\arctan(3x)=\frac{3}{1+9x^2}\), not \(\frac{1}{1+9x^2}\). The inner \(3x\) still contributes its derivative \(3\).

Swapping which one has the square root

\(\arcsin\) and \(\arccos\) give \(\pm\frac{1}{\sqrt{1-x^2}}\); \(\arctan\) gives \(\frac{1}{1+x^2}\). Keep the radical with the sine/cosine pair.

Losing the minus sign on \(\arccos\)

\(\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}\) — the exact negative of \(\arcsin\).

Evaluating \(f'\) at the wrong point

\(\big(f^{-1}\big)'(b)=\frac{1}{f'(a)}\) where \(a=f^{-1}(b)\). Find the matching input first; don’t plug \(b\) into \(f'\).

Writing \(\big(f^{-1}\big)'(x)=\frac{1}{f'(x)}\)

Missing the inner \(f^{-1}(x)\). The reciprocal of \(f'\) must be taken at the mirror input, not at \(x\).

On the AP exam

Inverse-value problems: find the matching input \(a=f^{-1}(b)\) first (often from a table), then \(\big(f^{-1}\big)'(b)=1/f'(a)\).
Memorize the trio: \(\arcsin\to\frac{1}{\sqrt{1-x^2}}\), \(\arccos\to\) its negative, \(\arctan\to\frac{1}{1+x^2}\).
If a trig function is still sitting in your answer, you haven’t finished — inverse-trig derivatives are algebraic.
Where \(f'(a)=0\), the inverse has a vertical tangent and isn’t differentiable — a favorite trap.

You can now differentiate composites, implicit relations, and inverses — the full toolkit. Next we turn the derivative on itself: differentiating again and again gives acceleration, concavity, and the patterns behind series. Higher-Order Derivatives is next — or head back to the Unit 3 Guide.