If $\log (x+y)=\log x y+3$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
If $\log (x+y)=\log x y+3$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
- $\left(\frac{y}{x}\right)^2$
- $-\left(\frac{x}{y}\right)^2$
- $-\left(\frac{y}{x}\right)^2$
- $\left(\frac{x}{y}\right)^2$
Solution
$\begin{aligned} & \log (x+y)=\log x y+3 \\ & \Rightarrow \log \left(\frac{x+y}{x y}\right)=3 \\ & \Rightarrow \frac{x+y}{x y}=e^3 \\ & \Rightarrow \frac{1}{y}+\frac{1}{x}=e^3 \\ & \text { Diff } \frac{-1}{y^2} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{1}{x^2}=0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y^2}{x^2}\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 1)
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