MHT CET · Maths · Differentiation
If \(x=e^{(y+e)^{(y+e)}(y+\ldots \ldots \infty)}\), then \(\frac{d y}{d x}=\)
- A \(\frac{1-x}{x}\)
- B \(\frac{1+x}{x}\)
- C \(\frac{1}{x}\)
- D \(\frac{x}{1+x}\)
Answer & Solution
Correct Answer
(A) \(\frac{1-x}{x}\)
Step-by-step Solution
Detailed explanation
Here \(x=e^{y+x}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(
\begin{array}{l}
1=e^{y+x}\left(\frac{d y}{d x}+1\right) \Rightarrow 1=e^{y+x} \frac{d y}{d x}+e^{y+x} \\\therefore 1=x \cdot \frac{d y}{d x}+x \\
\frac{d y}{d x}=\frac{1-x}{x}
\end{array}
\)
Differentiating w.r.t. \(\mathrm{x}\)
\(
\begin{array}{l}
1=e^{y+x}\left(\frac{d y}{d x}+1\right) \Rightarrow 1=e^{y+x} \frac{d y}{d x}+e^{y+x} \\\therefore 1=x \cdot \frac{d y}{d x}+x \\
\frac{d y}{d x}=\frac{1-x}{x}
\end{array}
\)
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