JEE Main 2023DifferentiationDifferentiationMediumQuestionSuppose f(x)=(2x+2−x)tanxtan−1(x2−x+1)(7x2+3x+1)3f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}f(x)=(7x2+3x+1)3(2x+2−x)tanxtan−1(x2−x+1). Then the value of f′(0)f^{\prime}(0)f′(0) is equal toOptionsAπ\piπBπ\sqrt{\pi}πC0Dπ2\frac{\pi}{2}2πCheck AnswerHide SolutionSolutionf′(0)=limh→0f(h)−f(0)h=limh→0(2h+2−h)tanhtan−1(h2−h+1)−0(7h2+3h+1)3h=π\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\ & =\sqrt{\pi} \end{aligned}f′(0)=h→0limhf(h)−f(0)=h→0lim(7h2+3h+1)3h(2h+2−h)tanhtan−1(h2−h+1)−0=π