3.7 Derivatives of Inverse Functions

Big idea

An inverse function undoes the original one, and its graph is the original graph reflected across the line $y = x$. Reflection swaps the roles of rise and run, so a piece of curve with slope $3$ becomes a piece of curve with slope $1/3$. That single geometric observation is the whole section: the derivative of an inverse is the reciprocal of the derivative of the original function, evaluated at the matching point.

The matching is the part that needs care. The point $(a, b)$ on the graph of $f$ becomes the point $(b, a)$ on the graph of the inverse. So to find the slope of the inverse at $x = b$, you need the slope of $f$ at $a$, and $a$ is the value the inverse sends $b$ to. The formula encodes that bookkeeping, which is why the inverse function appears inside $f'$ rather than outside it.

What makes the result useful is that it never asks you to solve for the inverse. Many one-to-one functions have inverses that cannot be written in closed form, and the formula sidesteps that entirely: if you can find one point the original function passes through, you can find the slope of its inverse at the reflected point.

Two families of derivatives fall out immediately. Rational powers, since $x^{1/n}$ is the inverse of $x^n$, which completes the proof of the power rule for all rational exponents. And the inverse trigonometric functions, whose derivatives turn out to be algebraic expressions with no trigonometry left in them - a surprise that makes them appear constantly later, when integration runs the process backwards.

Decoder

If $g$ is the inverse of a differentiable one-to-one function $f$, then $g'(x) = 1 / f'(g(x))$ at every $x$ where $f'(g(x))$ is not zero.

Read the right side from the inside out. First $g(x)$ locates the input that $f$ sends to $x$. Then $f'$ measures how fast $f$ is changing there. Then the reciprocal converts a rate of $f$ into a rate of $g$, because the two functions trade their input and output axes.

The exclusion matters. Where $f$ has a horizontal tangent, the reflected curve has a vertical one, and the inverse has no derivative there. The cube and cube root show it plainly: the cube is flat at the origin, so the cube root is vertical there, and $\frac{1}{3}x^{-2/3}$ is undefined at $x = 0$.

The classic mistake is writing $1/f'(x)$ instead of $1/f'(g(x))$. Those differ unless $f'$ happens to be constant. The reciprocal has to be taken at the point on the original curve, not at the point on the inverse curve, and $g(x)$ is what transports you between them.

Definitions and results

Inverse function derivative. Let $f$ be one-to-one and differentiable on an interval with inverse $g$. If $f'(g(x)) \ne 0$, then $g$ is differentiable at $x$ and

$$ g'(x) = \frac{1}{f'\big(g(x)\big)} $$

Leibniz form. Writing $y = g(x)$, so that $x = f(y)$, the statement becomes $\frac{dy}{dx} = 1 \big/ \frac{dx}{dy}$, with $\frac{dx}{dy}$ expressed in terms of $y$ and then converted back to $x$.

Where it comes from. Differentiating the identity $f(g(x)) = x$ with the chain rule gives $f'(g(x))\,g'(x) = 1$, and solving for $g'(x)$ is the formula. The chain rule is doing all the work.

Rational powers. Since $x^{1/n}$ inverts $x^n$, the formula gives $\frac{d}{dx}x^{1/n} = \frac{1}{n}x^{1/n - 1}$, and combining with the chain rule extends the power rule to every rational exponent $p/q$.

Inverse sine and cosine. For $-1 < x < 1$,

$$ \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1-x^2}} $$

Inverse tangent and cotangent. For all real $x$,

$$ \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}, \qquad \frac{d}{dx}\cot^{-1}x = \frac{-1}{1+x^2} $$

Inverse secant and cosecant. For $|x| > 1$,

$$ \frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}, \qquad \frac{d}{dx}\csc^{-1}x = \frac{-1}{|x|\sqrt{x^2-1}} $$

Pairs differ by a sign. Each co-function derivative is the negative of its partner, because the two functions in each pair add to a constant. Differentiating that constant sum gives zero, which is the whole explanation.

With the chain rule. For a differentiable inner function $u$, the versions you actually use read $\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1-u^2}}$ and $\frac{d}{dx}\tan^{-1}u = \frac{u'}{1+u^2}$, and similarly for the rest.

Worked examples

A slope without the inverse formula

Let $f(x) = x^3 + 2x + 1$. Its derivative $f'(x) = 3x^2 + 2$ is positive everywhere, so $f$ is increasing and one-to-one, and an inverse $g$ exists. Finding a formula for $g$ means solving a cubic, which is not worth doing.

Find $g'(4)$. First locate the matching point: $f(1) = 1 + 2 + 1 = 4$, so $g(4) = 1$. Then $f'(1) = 3 + 2 = 5$, and

$$ g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(1)} = \frac{1}{5} $$

The reflection check: $f$ passes through $(1,4)$ with slope $5$, so $g$ passes through $(4,1)$ with slope $1/5$.

Deriving the inverse sine formula

Let $y = \sin^{-1}x$, so that $\sin y = x$ with $y$ between $-\pi/2$ and $\pi/2$. Differentiate both sides with respect to $x$:

$$ \cos y \cdot \frac{dy}{dx} = 1 \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{1}{\cos y} $$

Now convert to $x$. From $\sin y = x$ and the Pythagorean identity, $\cos y = \pm\sqrt{1 - x^2}$, and on the interval where inverse sine lives the cosine is nonnegative, so the plus sign is the correct one:

$$ \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}} $$

The restricted range is what pins down the sign. The formula blows up as $x$ approaches $\pm 1$, which is right: sine is flat at those angles, so its inverse is vertical there.

Cube roots from cubes

Let $g(x) = x^{1/3}$, the inverse of $f(x) = x^3$. Since $f'(x) = 3x^2$,

$$ g'(x) = \frac{1}{3\big(x^{1/3}\big)^{2}} = \frac{1}{3x^{2/3}} = \frac{1}{3}x^{-2/3} $$

which is the power rule with exponent $1/3$. The formula fails at $x = 0$, where $f'(0) = 0$, and that is the vertical tangent of the cube root at the origin rather than a defect in the rule.

A product and a composition

Differentiate $y = x\tan^{-1}x$. The product rule with the inverse tangent derivative gives

$$ \frac{dy}{dx} = \tan^{-1}x + \frac{x}{1+x^2} $$

Now differentiate $y = \sin^{-1}(x^2)$, valid for $|x| \le 1$. The chain rule puts the inner derivative on top and the inner function inside the square root:

$$ \frac{dy}{dx} = \frac{2x}{\sqrt{1 - x^4}} $$

A check on the pattern: $\sin^{-1}x + \cos^{-1}x$ equals $\pi/2$ for every $x$ in $[-1,1]$, a constant. Adding the two derivative formulas gives $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$, exactly as a constant function requires.

Practice

Start with the general formula. These give you a function and a point and ask for the slope of the inverse, without ever asking you to invert anything.

Then the six inverse trigonometric derivatives, on their own and inside products, quotients and compositions. Watch the domains and the absolute value in the secant and cosecant formulas.

Quiz

Five items on the reciprocal slope formula for inverses and on derivatives of the inverse trigonometric functions.