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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 21 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 21 maths textbook solution

Answers (1)

Answer:

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

Hint:

Use chain rule

Given:

y=x \sin (a+y)

Solution:

y=x \sin (a+y)

Differentiate the given equation w.r.t x

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

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

\frac{d y}{d x}=x\left[\frac{d \sin (a+y)}{d(a+y)} \times \frac{d(a+y)}{d y} \times \frac{d y}{d x}\right]+\sin (a+y)                    [Using chain rule]

 

     =x\left[\cos (a+y) \times\left(\frac{d a}{d y}+\frac{d y}{d y}\right) \times \frac{d y}{d x}\right]+\sin (a+y)

     =x\left[\cos (a+y) \times(0+1) \times \frac{d y}{d x}\right]+\sin (a+y)

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

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

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

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

y=x \sin (a+y)

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

Put x=\frac{y}{\sin (a+y)}  in the above equation

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

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

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

Thus, proved

 

Posted by

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