#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 47 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$

Hint:

\begin{aligned} &\frac{d}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}}-1 \end{aligned}

Given:

$y=\cos ^{-1}\left\{\frac{2 x-3 \sqrt{1-x^{2}}}{\sqrt{13}}\right)$

Solution:

Let,$x=\cos \theta$

\begin{aligned} &\mathrm{y}=\cos ^{-1}\left\{\frac{2 \cos \theta-3 \sqrt{1-\cos ^{2} \theta}}{\sqrt{4}}\right\} \\ &\text { Using } \cos ^{2} \theta+\sin ^{2} \theta=1 \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sqrt{\sin ^{2} \theta}\right] \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sin \theta\right] \end{aligned}

$\cos \phi=\frac{2}{\sqrt{13}}$

Let,$\sin \phi=\sqrt{1-\cos ^{2}\phi}$

\begin{aligned} &=\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}+\sqrt{1-\frac{4}{13}} \\ &=\sqrt{\frac{13-4}{13}}=\sqrt{\frac{9}{13}} \end{aligned}

$\sin \phi=\frac{3}{\sqrt{13}}$

So,

\begin{aligned} &\begin{aligned} \mathrm{y} &=\cos ^{-1}\{\cos \phi \cos \theta-\sin \phi \sin \theta] \\ &=\cos ^{-1}[\cos (\theta+\phi)] \end{aligned} \\ &\text { Using, } \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned}

$\mathrm{y}=\phi+\theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos ^{-1}(\cos \theta)=\theta \text { if } 0 \varepsilon[0, \pi]\right]$

\begin{aligned} &\mathrm{y}=\cos ^{-1}\left(\frac{2}{\sqrt{13}}\right)+\cos ^{-1} \mathrm{x} \\ &\text { since, } \mathrm{x}=\cos \theta \\ &\cos \phi=\frac{2}{\sqrt{13}} \end{aligned}

Differentiating its with Respect to x,

\begin{aligned} &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}} \frac{d}{d x}\left(\frac{2}{\sqrt{13}}\right)+\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}} \times 0+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)} \\ &\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=0-\frac{1}{\sqrt{1-x^{2}}} \end{aligned}