Differentiating sin⁻¹ x and tan⁻¹ x

Undo the inverse first

$y={\sin }^{-1}x$y = sin⁻¹ x is the angle whose sine is x, taken between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2. So $y={\sin }^{-1}x$y = sin⁻¹ x says exactly the same as sin y = x, with y in that range. There is no rule yet for differentiating ${\sin }^{-1}$sin⁻¹ directly, but sin y = x can be differentiated, because sin is a function whose derivative is known.

All angles here are in radians. The derivative of sin is cos only in radians, and every result below depends on it.

Differentiate both sides

Differentiate sin y = x with respect to x. The right-hand side gives 1. On the left, y is a function of x, so the chain rule gives $\cos y\cdot \frac{dy}{dx}$cos y · dy/dx, as in implicit differentiation. So $\cos y\cdot \frac{dy}{dx}=1$cos y · dy/dx = 1, and dividing by cos y gives $\frac{dy}{dx}=\frac{1}{\cos y}$dy/dx = 1/cos y.

This is the inverse-function rule at work: the gradient of sin at the angle y is cos y, and the gradient of ${\sin }^{-1}$sin⁻¹ at the matching point x is its reciprocal.

Back to x

The answer should be in terms of x. From ${\sin }^{2}y+{\cos }^{2}y=1$sin²y + cos²y = 1, $\cos y=\pm \sqrt{1-{\sin }^{2}y}$cos y = ±√(1 − sin²y). Because y lies between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2, cos y is never negative, so the positive root is the right one. And sin y is x, so $\cos y=\sqrt{1-x^2}$cos y = √(1 − x²). Therefore $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$d/dx sin⁻¹ x = 1/√(1 − x²).

This holds for −1 < x < 1. At x = 1 and x = −1, cos y = 0 and there is nothing to divide by: the graph of ${\sin }^{-1}x$sin⁻¹ x stands vertical at its two ends. Near them the gradient grows without limit. At x = 0.99 the formula gives 7.088812, and the chord from 0.999 to 1 already has gradient 44.725087.

Check at two points. At x = 0 the formula gives $\frac{1}{\sqrt{1}}=1$1/√1 = 1, and the chord with h = 0.001 gives 1. At x = 0.5 it gives $\frac{1}{\sqrt{0.75}}$1/√0.75, about 1.154701, and the chord gives 1.155086.

${\cos }^{-1}x$cos⁻¹ x: the same moves, and a minus sign

$y={\cos }^{-1}x$y = cos⁻¹ x is the angle between 0 and $\pi$π whose cosine is x, so cos y = x. Differentiating both sides gives $-\sin y\cdot \frac{dy}{dx}=1$−sin y · dy/dx = 1, so $\frac{dy}{dx}=-\frac{1}{\sin y}$dy/dx = −1/sin y. For y between 0 and $\pi$π, sin y is never negative, so $\sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-x^2}$sin y = √(1 − cos²y) = √(1 − x²). Therefore $\frac{d}{dx}{\cos }^{-1}x=-\frac{1}{\sqrt{1-x^2}}$d/dx cos⁻¹ x = −1/√(1 − x²), again for −1 < x < 1.

There is a quicker way to see the minus sign. For every x in [−1, 1], ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$sin⁻¹ x + cos⁻¹ x = π/2, two complementary angles; at x = 0.3, for instance, the two add to 1.570796. A constant has derivative 0, so the two derivatives add to 0, and ${\cos }^{-1}x$cos⁻¹ x must have the negative of the gradient of ${\sin }^{-1}x$sin⁻¹ x. The graph of ${\cos }^{-1}x$cos⁻¹ x falls from $\pi$π to 0 as x goes from −1 to 1.

Check at two points. At x = 0 the formula gives −1, and the chord with h = 0.001 gives −1. At x = 0.6 it gives $-\frac{1}{\sqrt{0.64}}=-\frac{1}{0.8}=-1.25$−1/√0.64 = −1/0.8 = −1.25, and the chord gives −1.250587.

The curves $y={\sin }^{-1}x$y = sin⁻¹ x, rising from $-\frac{\pi }{2}$−π/2 to $\frac{\pi }{2}$π/2, and $y={\cos }^{-1}x$y = cos⁻¹ x, falling from $\pi$π to 0. The tangent to ${\sin }^{-1}x$sin⁻¹ x at x = 0.5 has gradient $\frac{1}{\sqrt{0.75}}$1/√0.75, about 1.155; the tangent to ${\cos }^{-1}x$cos⁻¹ x at x = 0.6 has gradient −1.25. Both curves stand vertical at x = −1 and x = 1.

${\tan }^{-1}x$tan⁻¹ x

$y={\tan }^{-1}x$y = tan⁻¹ x is the angle between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2 whose tangent is x, so tan y = x. The derivative of tan is ${\sec }^2$sec², so differentiating both sides gives ${\sec }^{2}y\cdot \frac{dy}{dx}=1$sec²y · dy/dx = 1, and $\frac{dy}{dx}=\frac{1}{{\sec }^{2}y}$dy/dx = 1/sec²y.

To get back to x, use $1+{\tan }^{2}y={\sec }^{2}y$1 + tan²y = sec²y. Since tan y = x, ${\sec }^{2}y=1+x^2$sec²y = 1 + x². Therefore $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$d/dx tan⁻¹ x = 1/(1 + x²). Here no root is taken and $1+x^2$1 + x² is never 0, so the formula holds for every x.

Check at two points. At x = 0 it gives 1, and the chord with h = 0.001 gives 1. At x = 1 it gives $\frac{1}{2}$1/2, and the chord gives 0.49975.

The graphs agree

The graph of ${\tan }^{-1}x$tan⁻¹ x climbs steepest at the origin, where its gradient is 1, and flattens out toward the levels $\frac{\pi }{2}$π/2 and $-\frac{\pi }{2}$−π/2 as x grows in either direction. Its derivative $\frac{1}{1+x^2}$1/(1 + x²) says the same: it is 1 at x = 0, $\frac{1}{2}$1/2 at x = 1, $\frac{1}{5}$1/5 at x = 2, and it falls toward 0 on both sides. At x = 2 the chord with h = 0.001 gives 0.19992.

Both derivatives in this lesson are algebraic: no trigonometric function is left in $\frac{1}{\sqrt{1-x^2}}$1/√(1 − x²) or $\frac{1}{1+x^2}$1/(1 + x²). That is why an integral of either shape has an inverse trigonometric answer.

The curve $y={\tan }^{-1}x$y = tan⁻¹ x and, softer, its derivative $y=\frac{1}{1+x^2}$y = 1/(1 + x²). At x = 1 the curve is at $\frac{\pi }{4}$π/4, about 0.785, and its tangent has gradient $\frac{1}{2}$1/2, the height of the derivative curve directly below.

With an inner function

When the inverse is applied to something other than x, the chain rule multiplies by the derivative of the inside. For $y={\sin }^{-1}\left(\frac{x}{2}\right)$y = sin⁻¹(x/2), the inside is $\frac{x}{2}$x/2, with derivative $\frac{1}{2}$1/2, so $\frac{dy}{dx}=\frac{1}{\sqrt{1-\frac{x^2}{4}}}\times \frac{1}{2}=\frac{1}{\sqrt{4-x^2}}$dy/dx = 1/√(1 − x²/4) × 1/2 = 1/√(4 − x²). At x = 1 that is $\frac{1}{\sqrt{3}}$1/√3, about 0.577350; the chord with h = 0.001 gives 0.577447.

For $y={\tan }^{-1}\left(2x\right)$y = tan⁻¹(2x), the inside 2x has derivative 2, so $\frac{dy}{dx}=\frac{2}{1+4x^2}$dy/dx = 2/(1 + 4x²). At x = 0.5 that is $\frac{2}{2}=1$2/2 = 1, and the chord gives 0.999001.

The usual mistakes

Reading ${\sin }^{-1}x$sin⁻¹ x as $\frac{1}{\sin x}$1/sin x. The index −1 here means the inverse function, the angle whose sine is x, not a reciprocal.

Differentiating forwards. ${\sec }^{2}x$sec²x is the derivative of tan x, and cos x is the derivative of sin x; the derivatives of the inverses are $\frac{1}{1+x^2}$1/(1 + x²) and $\frac{1}{\sqrt{1-x^2}}$1/√(1 − x²).

Mixing up the two forms. The root belongs to ${\sin }^{-1}$sin⁻¹ and ${\cos }^{-1}$cos⁻¹, with $1-x^2$1 − x² under it; ${\tan }^{-1}$tan⁻¹ has no root, and its identity adds: $1+x^2$1 + x².

Dropping the minus sign of ${\cos }^{-1}x$cos⁻¹ x, or putting one on ${\sin }^{-1}x$sin⁻¹ x. ${\sin }^{-1}x$sin⁻¹ x rises, so its gradient is positive; ${\cos }^{-1}x$cos⁻¹ x falls, so its gradient is negative.

A rocket

In the application below, an observer 800 m from a launch pad watches a rocket climb. The angle of elevation is ${\tan }^{-1}\left(\frac{h}{800}\right)$tan⁻¹(h/800), so its derivative, times the rocket's speed, is the rate the angle opens; turning the derivative over gives the meters of climb per radian.