Question

If \[y = x^x , \text{ find } \frac{dy}{dx} \text{ at } x = e\] ?

Answer 1

\[\text{ We have, y } = x^x ......... \left( i \right)\]

Taking log on both sides,

\[\log y = \log x^x \]

\[ \Rightarrow \log y = x \log x\]

\[\Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right)\]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left( \frac{1}{x} \right) + \log x \left( 1 \right)\]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 1 + \log x\]

\[ \Rightarrow \frac{dy}{dx} = y\left( 1 + \log x \right)\]

\[ \Rightarrow \frac{dy}{dx} = x^x \left( 1 + \log x \right) .............\left[\text{  using equation} \left( i \right) \right]\]

\[\text{ Puting x = e, we get}, \]

\[\frac{dy}{dx} = e^e \left( 1 + \log_e e \right)\]

\[ \Rightarrow \frac{dy}{dx} = e^e \left( 1 + 1 \right) .............\left[ \because \log_e e = 1 \right]\]

\[ \Rightarrow \frac{dy}{dx} = 2 e^e\]