Mathematics - Differentiation Question with Solution | TestHub

Given: $f\left(x\right)=\frac{\left(2^x+2^{-x}\right)\tan x\sqrt{{\tan }^{-1}\left(x^2-x+1\right)}}{{\left(7x^2+3x+1\right)}^3}$

We know that, $f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}$

$\Rightarrow f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{\frac{\left(2^h+2^{-h}\right)\tan h\sqrt{{\tan }^{-1}\left(h^2-h+1\right)}}{{\left(7h^2+3h+1\right)}^3}-\frac{\left(2^0+2^0\right)\tan 0\sqrt{{\tan }^{-1}\left(0-0+1\right)}}{{\left(0+0+1\right)}^3}}{h}$

$\Rightarrow f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{\left(2^h+2^{-h}\right)\tan h\sqrt{{\tan }^{-1}\left(h^2-h+1\right)}}{h{\left(7h^2+3h+1\right)}^3}$

$\Rightarrow f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{\left(2^h+2^{-h}\right)\sqrt{{\tan }^{-1}\left(h^2-h+1\right)}}{{\left(7h^2+3h+1\right)}^3}$

$\Rightarrow f^{\text{'}}\left(0\right)=\frac{\left(2^0+2^0\right)\sqrt{{\tan }^{-1}\left(1\right)}}{{\left(0+1\right)}^3}$

$\Rightarrow f^{\text{'}}\left(0\right)=\left(2\right)\sqrt{\frac{\pi }{4}}$

$\Rightarrow f^{\text{'}}\left(0\right)=\sqrt{\pi }$