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  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 19

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 19

Answers (1)

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Answer: \frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}}

Hint: You must know the rule of solving derivative of polynomial function.

Given: \sqrt{\frac{1+x}{1-x}}

Solution:

Let  y=\sqrt{\frac{1+x}{1-x}}

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

Differentiating with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}} \\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+x}{1-x}\right) \end{aligned}              [  using chain rule]

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

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

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} \times\left\{\frac{2}{(1-x)^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}} \end{aligned}

 

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