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  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 42

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 42

Answers (1)

Answer: \left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]

Hint: You must know about the rules of solving derivative of trigonometric function.

Given: y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}

Solution:

Let  y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}

Differentiate with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left[\frac{3 x^{2} \sin x}{\left(7-x^{2}\right)^{\frac{1}{2}}}\right]

\frac{dy}{dx}=\frac{\left(7-x^{2}\right)^{\frac{1}{2}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(3 x^{2} \sin x\right)-\left(3 x^{2} \sin x\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(7-x^{2}\right)^{\frac{1}{2}}}{\left[\left(7-x^{2}\right)^{\frac{1}{2}}\right]^{2}} \ldots \frac{d}{d x}\left(\frac{u}{v}\right)=   \frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}

\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{\frac{1}{2}-1}(-2 x)}{7-x^{2}}\right]

\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{-\frac{1}{2}}(-2 x)}{7-x^{2}}\right]

\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]

 

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