Q64 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
If $e^{-y} \cdot y=x$, then $\frac{d y}{d x}$ is
- A$\frac{y}{1-y}$
- B$\frac{1}{x y(1-y)}$
- C$\frac{1}{x(1-y)}$
- Correct answerD$\frac{\mathrm{y}}{\mathrm{x}(1-\mathrm{y})}$
Solution
$$
\begin{aligned}
& e^{-y} \cdot y=x \\
& \therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \ldots
\end{aligned}
$$
and
$$
\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}
$$
Now $y=\mathrm{xe}^{\mathrm{y}}$
$$
\begin{aligned}
& \therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \\
& \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1}
\end{aligned}
$$
From (1) and (2), we write
$$
\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}
$$
Hence, the correct answer is option: 'D'.