The Derivative of an Inverse Function

A curve and its mirror image

Take $f\left(x\right)=\frac{x^2}{2}$f(x) = x²/2 for $x\ge 0$x ≥ 0. Its inverse undoes it: if $y=\frac{x^2}{2}$y = x²/2 then $x=\sqrt{2y}$x = √(2y), so $f^{-1}\left(x\right)=\sqrt{2x}$f⁻¹(x) = √(2x). The graph of the inverse is the graph of f reflected in the line y = x, because reflecting in that line swaps the two coordinates of every point. The point (1.5, 1.125) is on $y=\frac{x^2}{2}$y = x²/2, and its mirror image (1.125, 1.5) is on $y=\sqrt{2x}$y = √(2x).

The two curves cross on the mirror line itself, at (0, 0) and (2, 2), where swapping x and y changes nothing.

Rise and run swap

At x = 1.5 the gradient of $y=\frac{x^2}{2}$y = x²/2 is f'(1.5) = 1.5. On its tangent, a run of 0.5 gives a rise of 0.75.

Reflect that little triangle in y = x. What went across now goes up, and what went up now goes across: a run of 0.75 and a rise of 0.5. So the tangent to the inverse at (1.125, 1.5) has gradient $0.5÷0.75=\frac{2}{3}$0.5 ÷ 0.75 = 2/3, which is 1 ÷ 1.5.

Check it on the inverse directly. From x = 1.125 to x = 1.126, $\sqrt{2x}$√(2x) rises from 1.5 to $\sqrt{2.252}$√2.252, and the chord gradient with h = 0.001 is 0.666519, just under $\frac{2}{3}$2/3.

The curve $y=\frac{x^2}{2}$y = x²/2, its inverse $y=\sqrt{2x}$y = √(2x), and the mirror line y = x between them. At (1.5, 1.125) the tangent to $y=\frac{x^2}{2}$y = x²/2 runs 0.5 and rises 0.75, a gradient of 1.5. At the mirror point (1.125, 1.5) the tangent to the inverse runs 0.75 and rises 0.5, a gradient of $\frac{2}{3}$2/3.

The rule

Name the matching points. If f(a) = b, then (a, b) is on the graph of f and (b, a) is on the graph of $f^{-1}$f⁻¹. The tangents at those two points are mirror images, so their gradients are reciprocals: ${\left(f^{-1}\right)}^{\prime }\left(b\right)=1/f^{\prime }\left(a\right)$(f⁻¹)'(b) = 1 / f'(a).

The derivative of f is read at a, the input, and the inverse's gradient belongs to b, the output. Reading f' at b gives the gradient of f somewhere else on the curve. For $y=\frac{x^2}{2}$y = x²/2, f'(1.125) is 1.125, and 1 ÷ 1.125, about 0.889, is the gradient of the inverse at $x\approx 0.633$x ≈ 0.633, not at x = 1.125.

The rule needs $f^{\prime }\left(a\right)\ne 0$f'(a) ≠ 0. Where the tangent to f is level, its mirror image is vertical, and a vertical line has no gradient. The rule also needs f to have an inverse at all near a, which a curve that keeps rising (or keeps falling) there does.

Why: the chain rule

Doing $f^{-1}$f⁻¹ and then f returns the input: $f\left(f^{-1}\left(x\right)\right)=x$f(f⁻¹(x)) = x for every x the inverse accepts. Differentiate both sides. The right-hand side gives 1. The left-hand side is f applied to the inside function $f^{-1}\left(x\right)$f⁻¹(x), so the chain rule gives $f^{\prime }\left(f^{-1}\left(x\right)\right)\times {\left(f^{-1}\right)}^{\prime }\left(x\right)$f'(f⁻¹(x)) × (f⁻¹)'(x).

So $f^{\prime }\left(f^{-1}\left(x\right)\right)\times {\left(f^{-1}\right)}^{\prime }\left(x\right)=1$f'(f⁻¹(x)) × (f⁻¹)'(x) = 1, and dividing gives ${\left(f^{-1}\right)}^{\prime }\left(x\right)=1/f^{\prime }\left(f^{-1}\left(x\right)\right)$(f⁻¹)'(x) = 1 / f'(f⁻¹(x)). At x = b, $f^{-1}\left(b\right)=a$f⁻¹(b) = a, and this is the rule again: ${\left(f^{-1}\right)}^{\prime }\left(b\right)=1/f^{\prime }\left(a\right)$(f⁻¹)'(b) = 1 / f'(a). The two rates multiply to 1, the way a rise over a run and a run over a rise do.

From two numbers

The rule needs only two facts about f. Given f(3) = 7 and f'(3) = 5, the input that gives 7 is 3, so ${\left(f^{-1}\right)}^{\prime }\left(7\right)=1/f^{\prime }\left(3\right)=\frac{1}{5}$(f⁻¹)'(7) = 1 / f'(3) = 1/5. There is no need to find a formula for the inverse.

One function with those two facts is $f\left(x\right)=x^2-x+1$f(x) = x² − x + 1: f(3) = 9 − 3 + 1 = 7, and f'(x) = 2x − 1 gives f'(3) = 5. For $x\ge \frac{1}{2}$x ≥ 1/2 its inverse can be written out: solving $y=x^2-x+1$y = x² − x + 1 for x gives $f^{-1}\left(x\right)=\frac{1+\sqrt{4x-3}}{2}$f⁻¹(x) = (1 + √(4x − 3))/2, and $f^{-1}\left(7\right)=\frac{1+5}{2}=3$f⁻¹(7) = (1 + 5)/2 = 3. Its derivative is $\frac{1}{\sqrt{4x-3}}$1/√(4x − 3), which at x = 7 is $\frac{1}{\sqrt{25}}=\frac{1}{5}$1/√25 = 1/5. The chord with h = 0.001 gives 0.199992.

Cube roots and logarithms

Take $f\left(x\right)=x^3$f(x) = x³, with f(2) = 8 and $f^{\prime }\left(2\right)=3\times 2^2=12$f'(2) = 3 × 2² = 12. Its inverse is the cube root, so the cube root has gradient $\frac{1}{12}$1/12 at 8. Directly, $\sqrt[3]{x}=x^{\frac{1}{3}}$∛x = x^(1/3) differentiates to $\left(\frac{1}{3}\right)x^{-\frac{2}{3}}$(1/3)x^(−2/3), and at 8 that is $\left(\frac{1}{3}\right)\times \left(\frac{1}{4}\right)=\frac{1}{12}$(1/3) × (1/4) = 1/12. The chord from 8 with h = 0.001 gives 0.08333.

The same step gives the derivative of ln x from that of $e^x$eˣ. $e^x$eˣ is its own derivative, so at x = 2 its gradient is $e^2$e². Its inverse, ln x, has gradient $1/e^2$1 / e² at the matching point $x=e^2$x = e², about 7.389, and $1/e^2$1 / e² is $\frac{1}{x}$1/x there. The same holds at every point, which is why ln x differentiates to $\frac{1}{x}$1/x. A centered chord at $e^2$e² gives 0.135335, the value of $\frac{1}{e^2}$1/e² to six places.

At 0 the cube has a level tangent: f'(0) = 0. The rule then has nothing to divide by, and the cube root stands vertical at the origin. Its chord from 0 to 0.001 already has gradient 100, and from 0 to 0.000001 it has gradient 10000.

The curve $y=x^3$y = x³ and its inverse $y=\sqrt[3]{x}$y = ∛x, mirror images in y = x. At a = 1.4 the point on the cube is (1.4, 2.744), with gradient $3\times 1.4^2=5.88$3 × 1.4² = 5.88; at the mirror point (2.744, 1.4) the cube root has gradient 1 ÷ 5.88, about 0.170. Drag the point along the cube: the two gradients always multiply to 1.

The usual mistakes

Keeping f' as it is. With f(2) = 9 and f'(2) = 4, the inverse has gradient $\frac{1}{4}$1/4 at 9, not 4.

Turning over the point instead of the gradient. With f(2) = 9 and f'(2) = 4, the answer is not $\frac{1}{9}$1/9: the reciprocal is taken of f', not of the place it is read.

Reading f' at the output. With f(4) = 2 and f'(4) = 6, ${\left(f^{-1}\right)}^{\prime }\left(2\right)$(f⁻¹)'(2) is $1/f^{\prime }\left(4\right)=\frac{1}{6}$1 / f'(4) = 1/6. The derivative of f at 2 belongs to a different point of the curve.

Treating $f^{-1}$f⁻¹ as $\frac{1}{f}$1/f. The inverse undoes f; it does not divide by it. The rule comes from $f\left(f^{-1}\left(x\right)\right)=x$f(f⁻¹(x)) = x, a composition, and not from a product.