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

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

Answers (1)

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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