Differentiate the following functions with respect to x :

$y=\tan ^{-1}\left(\frac{x-a}{x+a}\right)$

Dividing numerator and denominator by $x$

$\mathrm{y}=\tan ^{-1}\left(\frac{1-\frac{\mathrm{a}}{\mathrm{x}}}{1+1 \times \frac{\mathrm{a}}{\mathrm{x}}}\right)$

Using, $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

$y=\tan ^{-1} 1-\tan ^{-1} \frac{a}{x}$

Differentiating w.r.t $x$ we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1} 1-\tan ^{-1} \frac{a}{x}\right)$

$\frac{d y}{d x}=0-\frac{1}{1+\left(\frac{a}{x}\right)^{2}} \frac{d}{d x}\left(\frac{a}{x}\right)$

$\frac{d y}{d x}=-\frac{x^{2}}{a^{2}+x^{2}}\left(-\frac{a}{x^{2}}\right)$

$\frac{d y}{d x}=\frac{a}{a^{2}+x^{2}}$