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  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 20 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 20 maths

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Answer: \frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)

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

Given: \sin \left(\frac{1+x^{2}}{1-x^{2}}\right)

Solution:

Let      y=\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)

Differentiating with respect to x

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

\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right) \frac{d}{d x}\left(\frac{1+x^{2}}{1-x^{2}}\right)                  [  using chain rule]

\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)-\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)}{\left(1-x^{2}\right)^{2}}\right]...\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{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{\left(1-x^{2}\right)(2 x)-\left(1+x^{2}\right)(-2 x)}{\left(1-x^{2}\right)^{2}}\right]

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

\frac{d y}{d x}=\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)

 

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