Derivatives of Inverse Trigonometric Functions

Each inverse trigonometric function is the inverse of a trigonometric function restricted to an

interval on which that function is one-one, and its derivative can be found by writing the

inverse relation as an ordinary (implicit) equation and differentiating both sides -- exactly the

technique Section 5 develops in general, applied here to the six standard inverse functions.

Derivative of $\sin^{-1}x$. Let $y=\sin^{-1}x$, for $x\in(-1,1)$ and $y\in(-\pi/2,\pi/2)$;

this means $\sin y=x$. Differentiating both sides with respect to $x$ (using the chain rule on

the left, since $y$ is a function of $x$):

Since $y\in(-\pi/2,\pi/2)$, $\cos y>0$, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$ (taking the

positive root). Hence

Derivative of $\cos^{-1}x$. By an identical argument with $y=\cos^{-1}x$ ($\cos y=x$,

$y\in(0,\pi)$, so $\sin y>0$ and $\sin y=\sqrt{1-x^2}$): differentiating $\cos y=x$ gives

$-\sin y\,(dy/dx)=1$, so

(Note $\sin^{-1}x+\cos^{-1}x=\pi/2$ for all $x\in[-1,1]$ -- a constant -- so their derivatives

must be negatives of each other, exactly as found; this is a useful self-check.)

Derivative of $\tan^{-1}x$. Let $y=\tan^{-1}x$, so $\tan y=x$ for all real $x$, with

$y\in(-\pi/2,\pi/2)$. Differentiating: $\sec^2y\,(dy/dx)=1$, and using $\sec^2y=1+\tan^2y=1+x^2$,

Derivative of $\cot^{-1}x$. By the same method with $\cot y=x$: $-\text{cosec}^2y\,(dy/dx)=1$,

and $\text{cosec}^2y=1+\cot^2y=1+x^2$, so

Derivatives of $\sec^{-1}x$ and $\text{cosec}^{-1}x$ (for $|x|>1$), obtained the same way from

$\sec y=x$ and $\text{cosec}\,y=x$ respectively:

Combined with the chain rule. For a composite argument, e.g. $y=\sin^{-1}(x^2)$, treat $u=x^2$ …