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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 47

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 47

Answers (1)

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

Hint: To solve this equation we use uv'  form

Given:  x \sin (a+y)+\sin a \cdot \cos (a+y)=0

Solution:   (u v)^{\prime}=u^{\prime} v+v^{\prime} u

        \sin (a+y)+x \cdot \cos (a+y) y^{\prime}+\sin a\left(-\sin (a+y)\left(0+y^{\prime}\right)=0\right.

        y^{\prime}[x \cdot \cos (a+y)+(-\sin a) \cdot \sin (a+y)]=-\sin (a+y)

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

        \frac{d y}{d x}=\frac{-\sin ^{2}(a+y)}{-\sin a\left(\cos ^{2}(a+y)+\sin ^{2}(a+y)\right.} \quad\left(\sin ^{2} x+\cos ^{2} x=1\right)

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

Hence proved

Posted by

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