Question

The derivative of the function $f(x)=(\cos {)}^{-1}\left\{\frac{1}{\sqrt{13}}(2\cos x-3\sin x)\right\}+(\sin {)}^{-1}\left\{\frac{1}{\sqrt{13}}(2\cos x+3\sin x)\right\}$ with respect to $\sqrt{1+x^2}$ at $x=\frac{3}{4}$ is $\frac{a}{b}$ then the value of $a+b$ is equal to

Answer 1

Answer: 13

Given, $f\left(x\right)={\left(\cos \right)}^{-1}\left\{\frac{1}{\sqrt{13}}\left(2\cos \left(x\right)-3\sin \left(x\right)\right)+{\left(\sin \right)}^{-1}\left\{\frac{1}{\sqrt{13}}\left(2\cos \left(x\right)+3\sin \left(x\right)\right)\right\}\right\}$ or ${\left(\cos \right)}^{-1}\left\{\frac{2}{\sqrt{13}}\cos \left(x\right)-\frac{3}{\sqrt{13}}\sin \left(x\right)\right\}+{\left(\sin \right)}^{-1}\left\{\frac{2}{\sqrt{13}}\cos \left(x\right)+\frac{3}{\sqrt{13}}\sin \left(x\right)\right\}$
For simplifying $f\left(x\right)$ , let us assume $\cos \left(\theta \right)=\frac{2}{\sqrt{13}}$ and $\sin \left(\theta \right)=\frac{3}{\sqrt{13}}$ .
$f(x)={\left(cos\right)}^{-1}\left\{\cos \left(\theta \right)\cos \left(x\right)-\sin \left(\theta \right)\sin \left(x\right)\right\}+{\left(\sin \right)}^{-1}\left\{\cos \left(\theta \right)\cos \left(x\right)+\sin \left(\theta \right)\sin \left(x\right)\right\}$
As we know, $\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)=\cos \left(A+B\right)$ and
$\cos \left(A\right)\cos \left(B\right)+\sin \left(A\right)\sin \left(B\right)=\cos \left(A-B\right)$ .
$\therefore f\left(x\right)={\left(\cos \right)}^{-1}\left\{\cos \left(\theta +x\right)\right\}+{\left(\sin \right)}^{-1}\left\{\cos \left(\theta -x\right)\right\}$
As we know, ${\left(\sin \right)}^{-1}\left(A\right)=\frac{\pi }{2}-{\left(\cos \right)}^{-1}\left(A\right)$ .
$\therefore f\left(x\right)={\left(\cos \right)}^{-1}\left\{\cos \left(\theta +x\right)\right\}+\frac{\pi }{2}-{\left(\cos \right)}^{-1}\left\{\cos \left(\theta -x\right)\right\}$
Also, ${\left(\cos \right)}^{-1}\cos \left(A\right)=A.$
Hence, $f\left(x\right)=\left(\theta +x\right)+\frac{\pi }{2}-\left(\theta -x\right)$
$f\left(x\right)=\theta +x+\frac{\pi }{2}-\theta +x$
$f\left(x\right)=2x+\frac{\pi }{2}..$(i)
Let, $t=\sqrt{1+x^2}...\left(ii\right)$
We have to find the value of $\frac{d}{dt}f\left(x\right)$
As $f\left(x\right)$ is a function of $x$ and $t$ is also a function of $x$ ,
$\therefore \frac{d}{dt}\left(f\left(x\right)=\frac{\frac{d}{dx}\left\{f\left(x\right)\right\}}{\frac{dt}{dx}}\right)$
$\frac{d}{dt}f\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\times \frac{dx}{dt}\right)$ ......(iii)
Differentiating equation $\left(i\right)$ with respect to $x$ we get,
$\frac{d}{dx}\left\{f\left(x\right)\right\}=\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(\frac{\pi }{2}\right)$
$\frac{d}{dx}\left\{f\left(x\right)\right\}=2$
Also differentiating equation (ii) with respect to $x$ ,
$\frac{dt}{dx}=\frac{d}{dx}\left(\sqrt{1+x^2}\right)=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)$ $=\frac{2x}{2\sqrt{1+x^2}}=\frac{x}{\sqrt{1+x^2}}$
By equation (iii) , $\frac{d}{dt}\left(f\left(x\right)=2\times \frac{\sqrt{1+x^2}}{x}\right)$
at $x=\frac{3}{4}$ ,$\frac{d}{dt}\left(f\left(x\right)=\frac{2}{3}\times 4\sqrt{1+{\left(\frac{3}{4}\right)}^2}\right)$
$\frac{d}{dt}\left(f\left(x\right)=\frac{2}{3}\times 4\times \frac{5}{4}\right)$
$\frac{d}{dt}\left(f\left(x\right)=\frac{10}{3}\right)$