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

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