Limit Continuity And Differentiability Ques 58

Answer:

Correct Answer: 58.($\left(\frac{1}{2}+\frac{\pi}{4}\right)$)

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

Using first principle,

$ \begin{aligned} & f^{\prime}(1)=\lim _{h \rightarrow 0}\left[\frac{f(1+h)-f(1)}{h}\right] \\ & =\lim _{h \rightarrow 0}\left[\frac{(1+h) \tan ^{-1}(1+h)-\tan ^{-1}(1)}{h}\right] \\ & =\lim _{h \rightarrow 0}\left[\frac{\tan ^{-1}(1+h)-\tan ^{-1}(1)}{h}+\frac{h \tan ^{-1}(1+h)}{h}\right] \\ & =\lim _{h \rightarrow 0}\left[\frac{1}{h} \tan ^{-1}\left(\frac{h}{2+h}\right)+\tan ^{-1}(1+h)\right] \\ & =\lim _{h \rightarrow 0}\left[\frac{\tan ^{-1}\left(\frac{h}{2+h}\right)}{(2+h) \cdot \frac{h}{2+h}}\right]+\frac{\pi}{4} \\ & =\lim _{h \rightarrow 0} \frac{1}{2+h}\left(\frac{\tan ^{-1}\left(\frac{h}{2+h}\right)}{\frac{h}{(2+h)}}\right)+\frac{\pi}{4}=\frac{1}{2}+\frac{\pi}{4} \\ \end{aligned} $