Inverse Functions and Inverse Trigonometric Differentiation

If $g$ is the inverse of $f$, then $f(g(x)) = x$. Differentiating both sides by the chain rule, $f'(g(x))\cdot g'(x) = 1$, so:

• $g'(x) = \dfrac{1}{f'(g(x))}$ — the derivative of the inverse is the reciprocal of $f'$ evaluated at $g(x)$, not at $x$.
• To use it at a point $x = a$: first find $b = g(a)$ (the input that makes $f(b) = a$), then $g'(a) = \dfrac{1}{f'(b)}$.
• Geometrically: $f$ and $f^{-1}$ are **reflections across $y = x$**; the slope at a point and the slope at its mirror image are reciprocals.

Learn these six cold — they are reflexes:

• $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$, $\dfrac{d}{dx}\cos^{-1}x = -\dfrac{1}{\sqrt{1-x^2}}$
• $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$, $\dfrac{d}{dx}\cot^{-1}x = -\dfrac{1}{1+x^2}$
• $\dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{|x|\sqrt{x^2-1}}$, $\dfrac{d}{dx}\operatorname{cosec}^{-1}x = -\dfrac{1}{|x|\sqrt{x^2-1}}$

For an inner function, chain through: $\dfrac{d}{dx}\sin^{-1}(u) = \dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}$, and similarly for the rest. A handy identity for direct work: $\sec(\tan^{-1}x) = \sqrt{1+x^2}$ and $\tan(\sec^{-1}x) = \sqrt{x^2-1}$ — drawing the right triangle reads these off instantly.

Pick the substitution that matches the argument's shape, then read off the standard collapse:

• $x = \tan\theta$ turns: $\dfrac{2x}{1+x^2} = \sin 2\theta$; $\dfrac{1-x^2}{1+x^2} = \cos 2\theta$; $\dfrac{2x}{1-x^2} = \tan 2\theta$; $\dfrac{3x-x^3}{1-3x^2} = \tan 3\theta$.
• $x = \sin\theta$ turns: $3x - 4x^3 = \sin 3\theta$; $2x\sqrt{1-x^2} = \sin 2\theta$.
• $x = \cos\theta$ turns: $4x^3 - 3x = \cos 3\theta$; and the half-angle $\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \tan\dfrac{\theta}{2}$.

After substituting, $\sin^{-1}(\sin 3\theta) = 3\theta$ etc., so e.g. $\sin^{-1}(3x-4x^3) = 3\sin^{-1}x$ and $\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x$. Branch care: $\sin^{-1}(\sin\alpha) = \alpha$ only when $\alpha$ lies in the principal range $[-\pi/2, \pi/2]$; outside it the collapse picks up a sign or a $\pi - \alpha$ correction.

Two identities do almost all the work here:

• Arctan addition/subtraction: $\tan^{-1}x \pm \tan^{-1}y = \tan^{-1}\!\dfrac{x \pm y}{1 \mp xy}$ (subject to range/branch). Run it BACKWARDS: a fraction of the shape $\dfrac{a + x}{1 - ax}$ splits as $\tan^{-1}a + \tan^{-1}x$.
• Complementary identity: $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ (constant!), and likewise $\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$ and $\sec^{-1}x + \operatorname{cosec}^{-1}x = \dfrac{\pi}{2}$.

Because a constant has zero derivative, recognising these saves the entire calculation — the answer to 'differentiate $\sin^{-1}x + \cos^{-1}x$' is simply $0$.

To find $\dfrac{du}{dv}$ where $u$ and $v$ are both inverse-trig in $x$:

• Choose ONE substitution ($x = \sin\theta$, $\tan\theta$, or $\cos\theta$) that collapses both.
• Suppose it gives $u = a\theta$ and $v = b\theta$ (each a constant times the same angle).
• Then $\dfrac{du}{dv} = \dfrac{du/d\theta}{dv/d\theta} = \dfrac{a}{b}$ — the $d\theta$ cancels, so it is simply the ratio of the angle-multiples.

This works because both $u$ and $v$ become linear in $\theta$ after the collapse; the derivative of a constant-times-$\theta$ is just that constant.

For $h(x) = e^{g(x)}$ with $g$ an inverse-trig function:

• $h'(x) = e^{g(x)}\cdot g'(x) = h(x)\,g'(x)$, so the logarithmic-derivative ratio is $\dfrac{h'(x)}{h(x)} = g'(x)$.
• For $h(x) = e^{\sin^{-1}x}$: $\dfrac{h'}{h} = \dfrac{1}{\sqrt{1-x^2}}$. For $h(x) = e^{\cos^{-1}x}$: $\dfrac{h'}{h} = -\dfrac{1}{\sqrt{1-x^2}}$.
• A related monotonicity shape: $g(u) = 2\tan^{-1}(e^u) - \dfrac{\pi}{2}$ has $g'(u) = \dfrac{2e^u}{1+e^{2u}} > 0$ for all $u$, so it is strictly increasing, and $g(-u) = -g(u)$, so it is odd.