MHT CET · Maths · Differentiation
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
- A \(-4\)
- B \(-8\)
- C 4
- D 8
Answer & Solution
Correct Answer
(A) \(-4\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \log (x+y)=\log (x y)+a \\ & \Rightarrow \log \left(\frac{x+y}{x y}\right)=a \\ & \Rightarrow \frac{x+y}{x y}=e^a \\ & \Rightarrow \frac{1}{y}+\frac{1}{x}=e^a \\ & \Rightarrow-\frac{1}{y^2} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{1}{x^2}=0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y^2}{x^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \quad=-\frac{4^2}{2^2}=-4\end{aligned}\)
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