MHT CET · Maths · Differentiation
If \(e^x+e^y=e^{x+y}\), then \(\frac{d y}{d x}=\)
- A \(-e^{y-x}\)
- B \(e^{x-y}\)
- C \(-e^{x-y}\)
- D \(e^{y-x}\)
Answer & Solution
Correct Answer
(A) \(-e^{y-x}\)
Step-by-step Solution
Detailed explanation
\(e^x+e^y=e^{x+y} \Rightarrow e^x+e^y \cdot \frac{d y}{d x}=e^{x+y}\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{e^x-e^{x+y}}{e^{x+y}-e^y}=\frac{e^x-e^x-e^y}{e^x+e^y-e^y}=-\frac{e^y}{e^x}=-e^{y-x}\)
\(\Rightarrow \frac{d y}{d x}=\frac{e^x-e^{x+y}}{e^{x+y}-e^y}=\frac{e^x-e^x-e^y}{e^x+e^y-e^y}=-\frac{e^y}{e^x}=-e^{y-x}\)
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