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  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 24 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 24 maths

Answers (1)

Answer:

\left(x^{2}+1\right) \frac{d y}{d x}+x y+1=0

Hint:

Use product rule and chain rule

Given:

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

Solution:

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

Differentiate  w.r.t x

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

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

                                                                                                                    [Use product rule and chain rule]

\Rightarrow y \cdot\left[\frac{d\left(\sqrt{x^{2}+1}\right)}{d\left(x^{2}+1\right)} \times \frac{d\left(x^{2}+1\right)}{d x}\right]+\sqrt{x^{2}+1} \frac{d y}{d x}=\frac{1}{\left(\sqrt{x^{2}+1}-x\right)} \times\left(\frac{\left(\sqrt{x^{2}+1}\right)}{d x}-\frac{d x}{d x}\right)

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

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

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

    \frac{x y}{\sqrt{x^{2}+1}}+\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=-\frac{1}{\sqrt{x^{2}+1}}

    \sqrt{x^{2}+1} \cdot \frac{d y}{d x}=\frac{-1}{\sqrt{x^{2}+1}}-\frac{x y}{\sqrt{x^{2}+1}}

    \sqrt{x^{2}+1} \cdot \frac{d y}{d x}=\frac{-1-x y}{\sqrt{x^{2}+1}}

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

    \left(x^{2}+1\right) \frac{d y}{d x}+1+x y=0

Thus, proved

 

Posted by

infoexpert26

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