MCQ
If $x=e^{\left(\frac{x}{y}\right)}$, then $\frac{d y}{d x}=$
- A$\frac{x-y}{x \log y}$
- B$\frac{x-y}{x \log x}$
- C$\frac{x-y}{y \log x}$
- D$\frac{x+y}{x \log x}$
(b) : We have, $x=e^{x / y}$
By taking $\log$ on both sides, we get $\log x=\log e^{x / y}$
$\log x=\frac{x}{y}$ or $y \log x=x$
Differentiating w.r.t $x$ ',
$
\frac{d y}{d x} \log x+\frac{y}{x}=1 \Rightarrow \frac{d y}{d x}=\frac{1-\frac{y}{x}}{\log x}=\frac{x-y}{x \log x}
$
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