Question

Find the differential coefficient of ${\tan }^{-1}x$

Answer 1

Let $f\left(x\right)={\tan }^{-1}x$ and $f(x+\partial x)={\tan }^{-1}(x+\partial x)$

Now, $\frac{d}{dx}{\tan }^{-1}x=\underset{\partial x\to 0}{lim}\frac{{\tan }^{-1}(x+\partial x)-{\tan }^{-1}x}{\partial x}$

Let $t={\tan }^{-1}x\Rightarrow \tan t=x$

$t+\partial t={\tan }^{-1}(x+\partial x)$

On putting values

if $\partial x\to 0$ and $\partial t\to 0$

$\frac{d}{dx}{\tan }^{-1}x=\underset{\partial x\to 0}{lim}\frac{t+\partial t-t}{\tan (t+\partial t)-\tan t}$

$=\underset{\partial t\to 0}{lim}\frac{\partial t}{\frac{\sin (t+\partial t)}{\cos (t+\partial t)}-\frac{\sin t}{\cos t}}$

$=\underset{\partial t\to 0}{lim}\frac{\partial t\left(\cos t\cos (t+\partial t)\right)}{\cos t\sin (t+\partial t)-\sin t\cos (t+\partial t)}$

$=\underset{\partial t\to 0}{lim}\frac{\partial t[\cos t.\cos (t+\partial t)]}{\sin (t+\partial t-t)}$

$\underset{\partial t\to 0}{lim}\left(\frac{\partial t}{\sin \left(\partial t\right)}\right)\cos t.\cos (t+\partial t)=\left(1\right)\cos t\cos t$

$=\frac{1}{{\sec }^{2}t}$

$=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}$