Inverse Trig Derivatives PYQs

Let $x = \sin \theta$. Then $\sqrt{1-x^2} = \cos \theta$.

Function becomes $\sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta) = 2\theta$.

Substituting back: $2\sin^{-1} x$.

Derivative is $\frac{d}{dx}(2\sin^{-1} x) = \frac{2}{\sqrt{1-x^2}}$.

Let $x = \tan \theta$.

$y = \tan^{-1}\left(\frac{\tan(\pi/4) + \tan\theta}{1 - \tan(\pi/4)\tan\theta}\right) = \tan^{-1}(\tan(\pi/4 + \theta))$.

$y = \pi/4 + \theta = \pi/4 + \tan^{-1} x$.

$\frac{dy}{dx} = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.

$\frac{\sin x}{1+\cos x} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \tan(x/2)$.

So $y = \tan^{-1}(\tan(x/2)) = x/2$.

$\frac{dy}{dx} = 1/2$.

Divide numerator and denominator by $\cos x$:

$\tan^{-1}\left[\frac{1 - \tan x}{1 + \tan x}\right] = \tan^{-1}[\tan(\pi/4 - x)]$.

$= \pi/4 - x$.

Derivative is $-1$.

Let $x = \tan \theta \Rightarrow \theta = \tan^{-1} x$.

$\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \frac{\sec\theta-1}{\tan\theta} = \frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan(\theta/2)$.

So $y = \tan^{-1}(\tan(\theta/2)) = \theta/2 = \frac{1}{2}\tan^{-1} x$.

Derivative is $\frac{1}{2(1+x^2)}$.

Let $u = \sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) = \cos^{-1}\sqrt{1-x^2}$.

Let $x = \sin \theta \Rightarrow u = \cos^{-1}(\cos\theta) = \theta = \sin^{-1} x$.

$\frac{du}{dx} = \frac{1}{\sqrt{1-x^2}}$.

Let $v = \sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x$.

$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$.

$\frac{du}{dv} = \frac{1/\sqrt{1-x^2}}{2/\sqrt{1-x^2}} = 1/2$.

$\frac{1+\cos x}{\sin x} = \frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} = \cot(x/2) = \tan(\pi/2 - x/2)$.

So $y = \tan^{-1}(\tan(\pi/2 - x/2)) = \pi/2 - x/2$.

Derivative is $-1/2$.

Divide numerator and denominator by $b \cos x$:

$\tan^{-1}\left[\frac{(a/b) - \tan x}{1 + (a/b)\tan x}\right] = \tan^{-1}(a/b) - \tan^{-1}(\tan x)$.

$= \tan^{-1}(a/b) - x$.

Derivative is $0 - 1 = -1$.