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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)

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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