Question:
Differentiate $\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$ with respect to $\sqrt{1+a^{2} x^{2}}$.
Solution:
Let $u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$ and $v=\sqrt{1+a^{2} x^{2}}$
We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$.
We have $u=\tan ^{-1}\left(\frac{1+\mathrm{ax}}{1-\mathrm{ax}}\right)$
By substituting $a x=\tan \theta$, we have
$\mathrm{u}=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)$
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