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  • provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 22

provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 22

Answers (1)

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Answer:

\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}

Hint:

Use chain rule

Given:

x \sin (a+y)+\sin a \cos (a+y)=0

Solution:

\begin{aligned} &x \sin (a+y)+\sin a \cos (a+y)=0 \\ &x \sin (a+y)=-\sin a \cos (a+y) \end{aligned}

-\frac{x}{\sin a}=\frac{\cos (a+y)}{\sin (a+y)}

-\frac{x}{\sin a}=\cot (a+y)

Differentiating  equation w.r.t x

\frac{d}{d x}\left(\frac{-x}{\sin a}\right)=\frac{d}{d x} \cot (a+y)

\frac{-1}{\sin a} \cdot \frac{d x}{d x}=\frac{d \cot (a+y)}{d(a+y)} \times \frac{d(a+y)}{d x}                [Use chain rule]

\frac{-1}{\sin a}=-\cos e c^{2}(a+y) \times\left(\frac{d a}{d x}+\frac{d y}{d x}\right)            \left[\because \frac{d(\cot \theta)}{d \theta}=-\cos e c^{2} \theta\right]

-\frac{1}{\sin a}=-\cos e c^{2}(a+y)\left[0+\frac{d y}{d x}\right]

-\frac{1}{\sin a}=-\operatorname{cosec}^{2}(a+y) \cdot \frac{d y}{d x}

\frac{d y}{d x}=\frac{1}{\sin a \cdot \cos e c^{2}(a+y)}

\frac{d y}{d x}=\frac{1}{\sin a \cdot\left(\frac{1}{\sin ^{2}(a+y)}\right)}                            \left[\because \cos e c \theta=\frac{1}{\sin \theta}\right]

\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}

Thus, proved

 

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