Mathematics - Differentiation Question with Solution | TestHub

Given $f\left(x\right)=\sin \left({\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)\text{ at }x=1;2^{2x}={\left(2^x\right)}^2$

For $\sin \left({\cos }^{-1}\left(\frac{1-{\left(2^x\right)}^2}{1+{\left(2^x\right)}^2}\right)\right)$

Let ${\tan }^{-1}{2}^x=\theta ;\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

$\therefore \sin \left({\cos }^{-1}\cos 2\theta \right)=\sin 2\theta$

$\left\{\begin{array}{lr}\text{ If }& \mathrm{x}>1\Rightarrow \frac{\pi }{2}>\theta >\frac{\pi }{4}\\ \therefore & \pi >2\theta >\frac{\pi }{2}\end{array}\right\}$

$=2\sin \theta \cos \theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$

Hence, $f\left(x\right)=\frac{2·2^x}{1+2^{2x}}$

$\therefore f^{\text{'}}\left(x\right)=\frac{\left(1+2^{2x}\right)\left(2.2^x\ln 2\right)-2^{2x}·2·\ln 2·2·2^x}{\left(1+2^{2x}\right)}$

$\therefore f^{\text{'}}\left(1\right)=\frac{20\ln 2-32\ln 2}{25}=-\frac{12}{25}\ln 2$

So, $a=25, b=12$.

Then, $\left|a^2-b^2\right|={25}^2-{12}^2$

$=625-144$

$=481$