If $\log (x+y)=\log (x y)+a$, where a is constant, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=2$ and $y=4$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \log (x+y)=\log (x y)+a \ & \Rightarrow \log \left(\frac{x+y}{x y}ight)=a \ & \Rightarrow \frac{x+y}{x y}=e^a \ & \Rightarrow \frac{1}{y}+\frac{1}{x}=e^a \ & \Rightarrow-\frac{1}{y^2} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{1}{x^2}=0 \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y^2}{x^2} \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \quad=-\frac{4^2}{2^2}=-4\end{aligned}$

If $\log (x+y)=\log (x y)+a$, where a is constant, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=2$ and…

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