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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 58

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 58

Answers (1)

Answer:  y \cdot \frac{d y}{d x}+x=2 y

Hint:  To solve this we add log on both side

Given:  (x-y)^{e \frac{x}{x-y}}=\mathrm{a}

Solution:  

        e^{\frac{x}{x-y}}=\frac{a}{x-y}

Taking log on both sides,

        \log e^{\frac{x}{x-y}}=\log \left(\frac{a}{x-y}\right)

        \frac{x}{x-y} \log e=\log a-\log (x-y) \quad\left[\because \log \frac{p}{q}=\log p-\log q\right]

        \frac{x}{x-y}=\log a-\log (x-y)

Differentiate w.r.t x  we get

        \frac{d}{d x}\left(\frac{x}{x-y}\right)=0-\frac{d y}{d x} \log (x-y)

        \frac{\left(\frac{x}{x-y}\right) \frac{d x}{d x}-x^{\frac{d y}{d x}(x-y)}}{(x-y)^{2}}=\frac{-d}{d x} \log t                            \left[\because \frac{d}{d x} \frac{p(x)}{q(x)}=q(x) \frac{d p}{d x}-\frac{p \frac{d q}{d x}}{q^{2}}\right]

        \frac{(x-y) 1-x\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=\frac{-d}{d x} \log \frac{d t}{d x}                    \mathrm{x}-\mathrm{y}=\mathrm{t}=>\frac{d t}{d x}=\frac{d x}{d x}-\frac{d y}{d x}

                                                                                                                    \frac{d t}{d x}=1-\frac{d y}{d x}

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

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

        \begin{aligned} &=x \cdot \frac{d y}{d x}-x-y \frac{d y}{d x}+y \\\\ &-y=x+y-y \frac{d y}{d x} \\\\ &y \frac{d y}{d x}+x=2 y \end{aligned}

Hence proved

 

 

        

Posted by

infoexpert26

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