Advertisements
Advertisements
Question
Differentiate \[\left( x^x \right) \sqrt{x}\] ?
Advertisements
Solution
\[\text{ Let y } = x^x \sqrt{x} . . . \left( i \right)\]
\[\text{ Taking log on both sides }, \]
\[\log y = \log\left( x^x \sqrt{x} \right)\]
\[ \Rightarrow \log y = \log x^x + \log x^\frac{1}{2} \]
\[ \Rightarrow \log y = x \log x + \frac{1}{2}\log x \]
Differentiating with respect to x,
\[\frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right) + \frac{1}{2}\frac{d}{dx}\left( \log x \right) \]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left( \frac{1}{x} \right) + \log x\left( 1 \right) + \frac{1}{2}\left( \frac{1}{x} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 1 + \log x + \frac{1}{2x}\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ 1 + \log x + \frac{1}{2x} \right]\]
\[ \Rightarrow \frac{dy}{dx} = x^x \sqrt{x}\left[ 1 + \log x + \frac{1}{2x} \right] \left[ \text{ using equation} \left( i \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = x^{x + \frac{1}{2}} \left[ \left( \frac{2x + 1}{2x} \right) + \log x \right]\]
