Question

If \[x^y = e^{x - y} ,\text{ then } \frac{dy}{dx}\] is __________ .

Options

• \[\frac{1 + x}{1 + \log x}\]

• \[\frac{1 - \log x}{1 + \log x}\]

• not defined

• \[\frac{\log x}{\left( 1 + \log x \right)^2}\]

Answer 1

\[\frac{\log x}{\left( 1 + \log x \right)^2}\]

 

\[\text{ We have,} x^y = e^{x - y} \]
\[\text{ Taking log on both sides we get }, \]
\[ \Rightarrow y \log x = \left( x - y \right) \log_e e\]
\[ \Rightarrow y \log x = x - y\]
\[ \Rightarrow y \log x + y = x\]
\[ \Rightarrow y\left( 1 + \log x \right) = x\]
\[ \Rightarrow y = \frac{x}{1 + \log x}\]

\[\Rightarrow \frac{dy}{dx} = \frac{\left( 1 + \log x \right) \times 1 - x \times \left( 0 + \frac{1}{x} \right)}{\left( 1 + \log x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1 + \log x - 1}{\left( 1 + \log x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\log x}{\left( 1 + \log x \right)^2}\]