Aspire Faculty ID #15189 · Topic: JAMIA MCA 2017 · Just now
JAMIA MCA 2017

If $y=\tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)$, then $\dfrac{dy}{dx}$ equals …

Solution

$y'=\dfrac{u'}{1+u^{2}}$, $u=\dfrac{1+x}{1-x}$. $u'=\dfrac{2}{(1-x)^2}$, $1+u^2=\dfrac{2(1+x^2)}{(1-x)^2}$. Hence $y'=\dfrac{1}{1+x^2}$.

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