Suppose [f(x)=(2^x+2^-x) x ^-1(x^2-x+1)(7 x^2+3 x+1)^3].

Question

Suppose $$f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}$$. Then the value of $$f^{\prime}(0)$$ is equal to

Accepted Answer

$$\begin{aligned}
& f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \
& =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \
& =\sqrt{\pi}
\end{aligned}$$

Solutions · 1