How to Differentiate Inverse Functions

Inverse functions reverse what the original function did. If \(f\) maps \(x\) to \(y\), then \(f^{-1}\) maps \(y\) back to \(x\). Differentiating an inverse function requires a bit more care than differentiating the original — but there's a clean formula that makes it systematic, and the inverse trig functions give it its most important applications.

Inverse Function Review

A function \(f\) has an inverse on an interval if it is one-to-one there — meaning no two different inputs produce the same output. If two x-values mapped to the same y-value, then the inverse (working backwards) would need to map that y to two different x-values, which would make the inverse not a function.

If \(f\) is invertible, we write its inverse as \(f^{-1}\), and it satisfies:

$$f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$$

Each composition returns the original input. This two-way relationship is the key property we'll use to find the derivative.

Example: \(f(x) = x^3\) and \(f^{-1}(x) = x^{1/3}\) are inverses, since \((x^{1/3})^3 = x\) and \((x^3)^{1/3} = x\).

Example: \(\sin(x)\) and \(\arcsin(x)\) are inverses on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), since that's the interval where sine is one-to-one.

The Derivative Formula

Here's how to derive the formula for the derivative of an inverse function. Start with the defining property:

$$f(f^{-1}(x)) = x$$

Differentiate both sides with respect to \(x\), using the chain rule on the left:

$$f'(f^{-1}(x)) \cdot \frac{d}{dx}\left[f^{-1}(x)\right] = 1$$

Solve for the derivative of \(f^{-1}\):

$$\frac{d}{dx}\left[f^{-1}(x)\right] = \frac{1}{f'(f^{-1}(x))}$$

This is the inverse function derivative formula. In words: the derivative of the inverse function at \(x\) equals one divided by the derivative of the original function evaluated at \(f^{-1}(x)\).

There's an equivalent way to write this that's sometimes easier to work with. If we let \(y = f^{-1}(x)\) (so that \(f(y) = x\)), then:

$$\frac{dy}{dx} = \frac{1}{f'(y)}$$

The derivative of the original function evaluated at y — not at x. You then express the result in terms of \(x\) using the relationship \(f(y) = x\).

Using the General Formula

Example 1

Let \(f(x) = x^3\), so \(f^{-1}(x) = x^{1/3}\). Find \(\frac{d}{dx}\left[x^{1/3}\right]\) using the inverse function formula.

We have \(f'(x) = 3x^2\). The formula gives:

$$\frac{d}{dx}\left[f^{-1}(x)\right] = \frac{1}{f'(f^{-1}(x))} = \frac{1}{3\left(x^{1/3}\right)^2} = \frac{1}{3x^{2/3}}$$

This matches what the power rule gives directly: \(\frac{d}{dx}\left[x^{1/3}\right] = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\). ✓

Example 2

Let \(f(x) = x^5 + 2x^3\). This function is one-to-one (strictly increasing for all \(x\)), so \(f^{-1}\) exists — though we can't write it in a simple closed form. Find \([f^{-1}]'(3)\).

Notice that \(f(1) = 1 + 2 = 3\), so \(f^{-1}(3) = 1\). The formula gives:

$$[f^{-1}]'(3) = \frac{1}{f'(f^{-1}(3))} = \frac{1}{f'(1)}$$

Since \(f'(x) = 5x^4 + 6x^2\), we get \(f'(1) = 5 + 6 = 11\). Therefore:

$$[f^{-1}]'(3) = \frac{1}{11}$$

This is a powerful technique: even when you can't write the inverse function explicitly, you can still find the derivative at specific points — as long as you can find the corresponding input to the original function.

Inverse Trig Derivatives

The most common application of the inverse function formula in calculus is finding the derivatives of the inverse trig functions. Here is the complete list:

Function	Derivative	Domain
\(\arcsin(x)\)	\(\dfrac{1}{\sqrt{1-x^2}}\)	\(-1 < x < 1\)
\(\arccos(x)\)	\(\dfrac{-1}{\sqrt{1-x^2}}\)	\(-1 < x < 1\)
\(\arctan(x)\)	\(\dfrac{1}{1+x^2}\)	all real \(x\)
\(\text{arccot}(x)\)	\(\dfrac{-1}{1+x^2}\)	all real \(x\)
\(\text{arcsec}(x)\)	\(\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}\)	\(\lvert x\rvert > 1\)
\(\text{arccsc}(x)\)	\(\dfrac{-1}{\lvert x\rvert\sqrt{x^2-1}}\)	\(\lvert x\rvert > 1\)

The three you'll use most often — and the ones most likely to appear on exams — are arcsin, arccos, and arctan. Notice that arcsin and arccos are negatives of each other, and arctan and arccot are negatives of each other. That's not a coincidence: \(\arcsin(x) + \arccos(x) = \frac{\pi}{2}\) for all \(x\) in the domain, so their derivatives must sum to zero.

Deriving arcsin, arctan, and arccos

Rather than accepting the table on faith, let's derive the three most important entries using the inverse function formula. This also shows the technique you'd use to reconstruct any entry you forget.

Derivative of arcsin(x)

Let \(y = \arcsin(x)\), which means \(\sin(y) = x\) with \(y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Differentiate \(\sin(y) = x\) implicitly with respect to \(x\):

$$\cos(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos(y)}$$

Now express \(\cos(y)\) in terms of \(x\). From the Pythagorean identity \(\sin^2(y) + \cos^2(y) = 1\):

$$\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2}$$

(We take the positive square root because \(y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), where cosine is non-negative.)

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

Derivative of arctan(x)

Let \(y = \arctan(x)\), which means \(\tan(y) = x\) with \(y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Differentiate \(\tan(y) = x\) implicitly:

$$\sec^2(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y)$$

To express this in terms of \(x\): from \(\tan(y) = x\), we know \(\sec^2(y) = 1 + \tan^2(y) = 1 + x^2\). So:

$$\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \frac{1}{1 + x^2}$$

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

Derivative of arccos(x)

The derivation for arccos follows identically to arcsin, except that cosine is now the starting function. Let \(y = \arccos(x)\), so \(\cos(y) = x\) with \(y \in [0, \pi]\).

Differentiate implicitly:

$$-\sin(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{-1}{\sin(y)}$$

From \(\cos(y) = x\): \(\sin(y) = \sqrt{1 - x^2}\) (positive because \(y \in [0, \pi]\)).

$$\frac{d}{dx}[\arccos(x)] = \frac{-1}{\sqrt{1-x^2}}$$

Exactly the negative of the arcsin derivative, as expected.

The Chain Rule with Inverse Trig

When the argument of an inverse trig function is something other than bare \(x\), apply the chain rule: differentiate the inverse trig function, then multiply by the derivative of the inner expression.

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

$$\frac{d}{dx}[\arctan(u)] = \frac{1}{1+u^2} \cdot u'$$

Example 3

$$\frac{d}{dx}[\arcsin(2x)] = \frac{1}{\sqrt{1-(2x)^2}} \cdot 2 = \frac{2}{\sqrt{1-4x^2}}$$

Example 4

$$\frac{d}{dx}[\arctan(x^2)] = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4}$$

Example 5

$$\frac{d}{dx}[\arctan(5x+1)] = \frac{1}{1+(5x+1)^2} \cdot 5 = \frac{5}{1+(5x+1)^2}$$

Worked Examples

Example 6: Product rule + inverse trig

Find \(\frac{d}{dx}[x \arctan(x)]\).

Use the product rule:

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

Example 7: Evaluating at a point

If \(g(x) = \arcsin(x)\), find \(g'!\left(\frac{1}{2}\right)\).

$$g'(x) = \frac{1}{\sqrt{1-x^2}}$$

$$g'!\left(\frac{1}{2}\right) = \frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

Example 8: Using the general formula at a point

Let \(f(x) = \sin(x)\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Find \([f^{-1}]'!\left(\frac{\sqrt{2}}{2}\right)\).

Here \(f^{-1} = \arcsin\), and \(\arcsin!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\). The general formula gives:

$$[f^{-1}]'!\left(\frac{\sqrt{2}}{2}\right) = \frac{1}{f'!\left(\frac{\pi}{4}\right)} = \frac{1}{\cos!\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

This agrees with the arcsin derivative formula: \(\frac{1}{\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^2}} = \frac{1}{\sqrt{1 - \frac{1}{2}}} = \frac{1}{\sqrt{\frac{1}{2}}} = \sqrt{2}\). ✓

Practice Problems

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

Show answerChain rule: \(\frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{1+9x^2}\)

Find \(\frac{d}{dx}[\arcsin(x^3)]\).

Show answerChain rule: \(\frac{1}{\sqrt{1-(x^3)^2}} \cdot 3x^2 = \frac{3x^2}{\sqrt{1-x^6}}\)

Find \(\frac{d}{dx}[x^2 \arctan(x)]\).

Show answerProduct rule: \(2x \cdot \arctan(x) + x^2 \cdot \frac{1}{1+x^2} = 2x\arctan(x) + \frac{x^2}{1+x^2}\)

Let \(f(x) = x^7\). Find \([f^{-1}]'(1)\).

Show answer\(f(1) = 1\) so \(f^{-1}(1) = 1\). \(f'(x) = 7x^6\), so \(f'(1) = 7\). Therefore \([f^{-1}]'(1) = \frac{1}{7}\).

Find \(\frac{d}{dx}[\arctan(\sqrt{x})]\).

Show answerChain rule with \(u = \sqrt{x}\), \(u' = \frac{1}{2\sqrt{x}}\): \(\frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1+x)}\)

Why does \(\frac{d}{dx}[\arcsin(x)] + \frac{d}{dx}[\arccos(x)] = 0\)?

Show answerBecause \(\arcsin(x) + \arccos(x) = \frac{\pi}{2}\) (a constant) for all \(x\) in \([-1,1]\). The derivative of a constant is zero, so the sum of the derivatives must be zero. This means \(\frac{d}{dx}[\arccos(x)] = -\frac{d}{dx}[\arcsin(x)] = \frac{-1}{\sqrt{1-x^2}}\).

Check your work with our Derivative Calculator.