MHT CET · Maths · Differentiation
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 \(\frac{1}{x(1-y)}\)
- D \(\frac{\mathrm{y}}{\mathrm{x}(1-\mathrm{y})}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{y}}{\mathrm{x}(1-\mathrm{y})}\)
Step-by-step Solution
Detailed explanation
\(e^{-y} \cdot y=x\)
\(\therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \ldots\)
and
\(
\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}
\)
Now \(y=\mathrm{xe}^{\mathrm{y}}\)
\(\therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \)
\( \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1}\)
From (1) and (2), we write
\(
\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}
\)
\(\therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \ldots\)
and
\(
\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}
\)
Now \(y=\mathrm{xe}^{\mathrm{y}}\)
\(\therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \)
\( \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1}\)
From (1) and (2), we write
\(
\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}
\)
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