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explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 20 maths

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


Use product rule and chain rule


x y \log (x+y)=1


x y \log (x+y)=1

Differentiate the given equation w.r.t x

\frac{d}{d x}[x y \log (x+y)]=\frac{d(1)}{d x}

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

x y\left[\frac{d \log (x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}\right]+\log (x+y)\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]=0           [Use product rule and chain rule] 

x y\left[\frac{1}{x+y} \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)\right]+\log (x+y)\left[x \frac{d y}{d x}+y\right]=0

x y\left[\frac{1}{x+y}+\frac{1}{x+y} \frac{d y}{d x}\right]+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0

\frac{x y}{x+y}+\frac{x y}{x+y} \frac{d y}{d x}+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0

\frac{d y}{d x}\left[\frac{x y}{x+y}+x \log (x+y)\right]=-\frac{x y}{x+y}-y \log (x+y)

Also   x y \log (x+y)=1                         \left[\log (x+y)=\frac{1}{x y}\right]

Put \log (x+y)=\frac{1}{x y}   in the above equation

\frac{d y}{d x}\left[\frac{x y}{x+y}+x \cdot \frac{1}{x y}\right]=-\frac{x y}{x+y}-y \cdot \frac{1}{x y}

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

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

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

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

Thus, proved

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