Question:
Differentiate the following functions with respect to $x$ :
$\cos ^{-1}\left(\frac{1-x^{2 n}}{1+x^{2 n}}\right), 0
Solution:
$y=\cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\}$
Let $x^{n}=\tan \theta$
Now
$y=\cos ^{-1}\left\{\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right\}$
Using $\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$
$y=\cos ^{-1}\{\cos 2 \theta\}$
Considering the limits,
$0 $0 $0<\theta<\frac{\pi}{2}$ Now, $y=\cos ^{-1}(\cos 2 \theta)$ $y=2 \theta$ $y=\tan ^{-1}\left(x^{n}\right)$ Differentiating w.r.t $\mathrm{x}$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(x^{n}\right)\right)$ $\frac{d y}{d x}=\frac{2 n x^{n-1}}{1+\left(x^{n}\right)^{2}}$ $\frac{d y}{d x}=\frac{2 n x^{n-1}}{1+x^{2 n}}$