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Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 54

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Answer: \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}

Hint: To solve this equation we solve side by side

Given: x y=e^{(x-y)}


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

                                                                        \left[\because \frac{d u v}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\right]

Diff w.r.t  x^{u}                                                       \left[\because \frac{d e^{x}}{d x}=e^{x}\right]

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

        \begin{aligned} &x \frac{d y}{d x}+y=e^{(x-y)}-e^{(x-y)} \frac{d y}{d x} \\\\ &x \frac{d y}{d x}+e^{(x-y)} \frac{d y}{d x}=e^{(x-y)}-y \end{aligned}

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


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