KCET · Maths · Application of Derivatives
If \((x e)^{y}=e^{y}\), then \(\frac{d y}{d x}\) is
- A \(\frac{\log x}{(1+\log x)^{2}}\)
- B \(\frac{1}{(1+\log x)^{2}}\)
- C \(\frac{\log x}{(1+\log x)}\)
- D \(\frac{e^{x}}{x(y-1)}\)
Answer & Solution
Correct Answer
(A) \(\frac{\log x}{(1+\log x)^{2}}\)
Step-by-step Solution
Detailed explanation
We have, \((x e)^{y}=e^{x}\)
Taking log on both sides at base \(e\), we get
\(\begin{aligned}
y \log (x e) &=x \log e \\
\Rightarrow \quad & y(\log x+\log e) &=x\left(\because \log _{e} e=1\right) \\
\Rightarrow \quad y &=\frac{x}{\log x+1}
\end{aligned}\)
On differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
\frac{d y}{d x} &=\frac{(\log x+1)-x\left(\frac{1}{x}+0\right)}{(\log x+1)^{2}} \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{\log x}{(\log x+1)^{2}}
\end{aligned}\)
Taking log on both sides at base \(e\), we get
\(\begin{aligned}
y \log (x e) &=x \log e \\
\Rightarrow \quad & y(\log x+\log e) &=x\left(\because \log _{e} e=1\right) \\
\Rightarrow \quad y &=\frac{x}{\log x+1}
\end{aligned}\)
On differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
\frac{d y}{d x} &=\frac{(\log x+1)-x\left(\frac{1}{x}+0\right)}{(\log x+1)^{2}} \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{\log x}{(\log x+1)^{2}}
\end{aligned}\)
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