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need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 23

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Answer:

\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}

Hint:

Use product rule and chain rule

Given:

y=x \sin y

Solution:

y=x \sin y

Differentiate w.r.t x

\frac{d y}{d x}=\frac{d(x \sin y)}{d x}

\frac{d y}{d x}=x \cdot \frac{d(\sin y)}{d x}+\sin y \cdot \frac{d x}{d x}                            [Using product rule]

\frac{d y}{d x}=x \cdot \frac{d \sin y}{d y} \cdot \frac{d y}{d x}+\sin y                            [Using chain rule]

\frac{d y}{d x}=x \cdot \cos y \cdot \frac{d y}{d x}+\sin y

\frac{d y}{d x}-x \cos y \frac{d y}{d x}=\sin y

\frac{d y}{d x}(1-x \cos y)=\sin y

\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}

Thus, proved

 

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