Question

Find \[\frac{dy}{dx}\] \[y = x^{\log x }+ \left( \log x \right)^x\] ?

Answer 1

\[\text{ Let y }= x^{\log x }+ \left( \log x \right)^x \]

\[\text{ Also, let u } = \left( \log x \right)^x \text{ and v} = x^{\log x} \]

\[ \therefore y = v + u\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} + \frac{du}{dx} . . . \left( i \right)\]

\[\text{ Now, u} = \left( \log x \right)^x \]

\[ \Rightarrow \log u = \log\left[ \left( \log x \right)^x \right]\]

\[ \Rightarrow \log u = x\log\left( \log x \right)\]

Differentiating both sides with respect to x,

\[\frac{1}{u}\frac{du}{dx} = \log\left( \log x \right)\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left[ \log\left( \log x \right) \right]\]

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

\[ \Rightarrow \frac{du}{dx} = \left( \log x \right)^x \left[ \log\left( \log x \right) + \frac{x}{\log x} \times \frac{1}{x} \right]\]

\[ \Rightarrow \frac{du}{dx} = \left( \log x \right)^x \left[ \log\left( \log x \right) + \frac{1}{\log x} \right] . . . \left( ii \right)\]

\[\text{ Also, v} = x^{\log x} \]

\[ \Rightarrow \log v = \log x^{\log x} \]

\[ \Rightarrow \log v = \log x \log x = \left( \log x \right)^2 \]

Differentiating both sides with respect to x,

\[\frac{1}{v}\frac{dv}{dx} = \frac{d}{dx}\left[ \left( \log x \right)^2 \right]\]

\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = 2\left( \log x \right)\frac{d}{dx}\left( \log x \right)\]

\[ \Rightarrow \frac{dv}{dx} = 2v\left( \log x \right)\frac{1}{x}\]

\[ \Rightarrow \frac{dv}{dx} = 2 x^{\log x} \frac{\log x}{x}\]

\[ \Rightarrow \frac{dv}{dx} = 2 x^{\log x} \frac{\log x}{x} . . . \left( iii \right)\]

\[\text{ From} \left( i \right), \left( ii \right) \text{ and }\left( iii \right), \text{ we obtain}\]

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