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  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 31

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 31

 

Answers (1)

best_answer

Answer:

\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}     At x=\frac{\pi}{4}

Hint:

Use chain rule

Given:

\sqrt{y+x}+\sqrt{y-x}=c

Solution:

\sqrt{y+x}+\sqrt{y-x}=c

Differentiate the given equation w.r.t x

\frac{d}{d x}(\sqrt{y+x}+\sqrt{y-x})=\frac{d(c)}{d x}

\frac{d(\sqrt{y+x})}{d x}+\frac{d(\sqrt{y-x})}{d x}=0                        \left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]

 

\frac{d(\sqrt{y+x})}{d(y+x)} \times \frac{d(y+x)}{d x}+\frac{d(\sqrt{y-x})}{d(y-x)} \times \frac{d(y-x)}{d x}=0            [Using chain rule]

 

\frac{1}{2 \sqrt{y+x}} \times\left(\frac{d y}{d x}+\frac{d x}{d x}\right)+\frac{1}{2 \sqrt{y-x}} \times\left(\frac{d y}{d x}-\frac{d x}{d x}\right)=0            \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

 

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

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

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

\frac{d y}{d x} \times[\sqrt{y-x}+\sqrt{y+x}]=[\sqrt{y+x}-\sqrt{y-x}]

\frac{d y}{d x}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}

 

Rationalizing the denominator

 

\frac{d y}{d x}=\frac{(\sqrt{y+x}-\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}{(\sqrt{y+x}+\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}

      =\frac{(\sqrt{y+x}-\sqrt{y-x})^{2}}{(\sqrt{y+x})^{2}-(\sqrt{y-x})^{2}}                \left[\because a^{2}-b^{2}=(a+b)(a-b)\right]

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

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

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

       =\frac{2(y-\sqrt{y+x} \cdot \sqrt{y-x})}{2 x}

       =\frac{y-\sqrt{y+x} \cdot \sqrt{y-x}}{x}

      =\frac{y-\sqrt{y^{2}-x^{2}}}{x}

      =\frac{y}{x}-\frac{\sqrt{y^{2}-x^{2}}}{\sqrt{x^{2}}}

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

Thus proved

 

 

Posted by

infoexpert26

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