Question

The derivative of the function, $f\left(x\right)={\cos }^{-1}\left\{\frac{1}{\sqrt{13}}(2\cos x-3\sin x)\right\}+{\sin }^{-1}\left\{\frac{1}{\sqrt{13}}(2\cos x+3\sin x)\right\}$
w.r.t $\sqrt{1+x^2}$ at $x=\frac{3}{4}$ is

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{10}{3}$
• D. $0$

Answer 1

Answer: (C)

$\frac{10}{3}$

Let $u=f\left(x\right)={\cos }^{-1}\left\{\frac{1}{\sqrt{3}}\right(2\cos x-3\sin x\left)\right\}+{\sin }^{-1}\left\{\frac{1}{\sqrt{13}}\right(2\cos x+3\sin x\left)\right\}$

Let $\cos \varphi =\frac{2}{\sqrt{3}}$

$\sin \varphi =\sqrt{1-{\cos }^2\varphi }$

$=\sqrt{1-\frac{4}{13}}$

$=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}$

$\therefore u={\cos }^{-1}\{\frac{2}{\sqrt{13}}\cos x-\frac{3}{\sqrt{13}}\}+{\sin }^{-1}\{\frac{2}{\sqrt{13}}\cos x+\frac{3}{\sqrt{13}}\sin x\}$

$={\cos }^{-1}\left\{\cos \right(\varphi +x\left)\right\}+{\sin }^{-1}\left\{\cos \right(\varphi -x\left)\right\}$

$=\varphi +x+\frac{\pi }{2}-{\cos }^{-1}\left\{\cos \right(\varphi -x\left)\right\}$

$=\varphi +x+\frac{\pi }{2}-\varphi +x$

$u=\frac{\pi }{2}+2x$

$\frac{du}{dx}=2$

Let $v=\sqrt{1+x^2}$

$\Rightarrow \frac{dv}{dx}=\frac{2x}{2\sqrt{1+x^2}}=\frac{x}{\sqrt{2+x^2}}$

$\frac{du}{dx}=2$

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{2}{\frac{2x}{2\sqrt{1+x^2}}}=\frac{2\sqrt{1+x^2}}{x}$

$\therefore {\begin{array}{c}\frac{du}{dv}\end{array}|}_{x=3/4}=\frac{2\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}=\frac{2\sqrt{\frac{25}{16}}}{\frac{3}{4}}=\frac{\frac{5}{4}}{\frac{3}{4}}=\frac{5}{3}\times 2$

$\therefore {\begin{array}{c}\frac{du}{dv}\end{array}|}_{x=3/4}=\frac{2\times 5}{3}=\frac{10}{3}$

$\therefore {\begin{array}{c}\frac{du}{dv}\end{array}|}_{x=3/4}=\frac{10}{3}$

$\to$ option C