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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)

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

 

Posted by

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