Derivatives of Standard Inverse Trigonometric Functions

This section derives the standard derivative of each inverse trigonometric function, by writing $y=(\text{inverse trig of }x)$ as $x=(\text{trig of }y)$, differentiating implicitly with respect to $y$, and converting the result back into $x$ using a Pythagorean identity — with the sign fixed by which quadrant the principal-branch angle $y$ can lie in.

1. $y=\sin^{-1}x$, $-1\le x\le1$, $-\frac{\pi}{2}\le y\le\frac{\pi}{2}$. Then $x=\sin y$. Differentiating w.r.t. $y$: $\dfrac{dx}{dy}=\cos y=\pm\sqrt{1-\sin^2y}=\pm\sqrt{1-x^2}$. Since $-\frac\pi2\le y\le\frac\pi2$, $y$ lies in the 1st or 4th quadrant, where $\cos y\ge0$, so the positive sign is taken: $\dfrac{dx}{dy}=\sqrt{1-x^2}$. By the inverse-function theorem, $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$, for $|x|<1$.

2. $y=\cos^{-1}x$, $0\le y\le\pi$: left as an exercise for the student to prove by the same method (writing $x=\cos y$); the standard result is $\dfrac{dy}{dx}=-\dfrac1{\sqrt{1-x^2}}$.

3. $y=\cot^{-1}x$, $x\in\mathbb R$, $0<y<\pi$. Then $x=\cot y$. Differentiating w.r.t. $y$: $\dfrac{dx}{dy}=-\text{cosec}^2y=-(1+\cot^2y)=-(1+x^2)$. So $\dfrac{dy}{dx}=\dfrac{1}{-(1+x^2)}=-\dfrac1{1+x^2}$.

4. $y=\tan^{-1}x$, $-\frac\pi2<y<\frac\pi2$: left as homework by the same method (writing $x=\tan y$, $\dfrac{dx}{dy}=\sec^2y=1+\tan^2y=1+x^2$); the standard result is $\dfrac{dy}{dx}=\dfrac{1}{1+x^2}$.

5. $y=\sec^{-1}x$, $|x|\ge1$, $0\le y\le\pi$, $y\ne\frac\pi2$. Then $x=\sec y$. Differentiating: $\dfrac{dx}{dy}=\sec y\tan y=\pm\sec y\sqrt{\sec^2y-1}=\pm x\sqrt{x^2-1}$. The sign is fixed by noting $\sec y$ and $\tan y$ are both positive in Quadrant I and both negative in Quadrant II, so their product $\sec y\tan y$ is positive throughout this branch; matching that against $x>0$ (Quadrant I, giving $x\sqrt{x^2-1}>0$) and $x<0$ (Quadrant II, giving $-x\sqrt{x^2-1}>0$) shows $\dfrac{dx}{dy}=x\sqrt{x^2-1}$ when $x>1$ and $\dfrac{dx}{dy}=-x\sqrt{x^2-1}$ when $x<-1$. Reciprocating: $\dfrac{dy}{dx}=\dfrac1{x\sqrt{x^2-1}}$ for $x>1$, and $\dfrac{dy}{dx}=-\dfrac1{x\sqrt{x^2-1}}$ for $x<-1$.

Note 1. A function is increasing where its derivative is positive and decreasing where its derivative is negative. Note 2. The derivative of $\sec^{-1}x$ is always positive, because the graph of $\sec^{-1}x$ is always increasing (on each piece of its domain).

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