Question

Find the differential coefficient of ${\tan }^{-1}x$ w.r to $x$.

Answer 1

Let $ta{n}^{-1}x=y$

$\therefore x=tany$

Now, differentiate wrt $x$

$\therefore 1=\frac{d\left(tany\right)}{dx}$ (differentiation of $tant$ is $se{c}^{2}t$)

$\therefore 1=se{c}^{2}y\frac{dy}{dx}$

$\therefore \frac{dy}{dx}=\frac{1}{se{c}^{2}y}$

$\therefore \frac{dy}{dx}=\frac{1}{1+ta{n}^{2}y}$ ($\because se{c}^{2}t=1+ta{n}^{2}t$)

But, $tany=x$

$\therefore \frac{d\left(ta{n}^{-1}x\right)}{dx}=\frac{1}{1+x^2}$

Hence the answer is $\frac{1}{1+x^2}$