Question

If f(x)=x tan^ -1 x, then f'(1) equals​

Answer 1

Answer:

Step-by-step explanation:

We have,

$f(x)=x\,\tan^{-1}(x)$

$\implies\,f^{\prime}(x)=\tan^{-1}(x)\cdot\dfrac{d}{dx}(x)+x\cdot\dfrac{d}{dx}\left(\tan^{-1}(x)\right)$

We know,

$\boxed{\dfrac{d}{dx}\Big\{\tan^{-1}(x)\Big\}=\dfrac{1}{1+x^2}}$

So,

$\implies\,f^{\prime}(x)=\tan^{-1}(x)\cdot1+\dfrac{x}{1+x^{2}}$

$\implies\,f^{\prime}(x)=\tan^{-1}(x)+\dfrac{x}{1+x^{2}}$

Now,

$\implies\,f^{\prime}(1)=\tan^{-1}(1)+\dfrac{1}{1+(1)^{2}}$

$\implies\,f^{\prime}(1)=\dfrac{\pi}{4}+\dfrac{1}{1+1}$

$\implies\,f^{\prime}(1)=\dfrac{\pi}{4}+\dfrac{1}{2}$