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

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

Answers (1)

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}

 

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