Question

If y = x^x, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]

Answer 1

\[\text { Given,} y = x^x \]

\[\text { Taking logarithm on both sides, we get }\]

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

\[\text { Differentiating both sides w . r . t . x, we get }\]

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

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

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

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

\[\text { Now }, \frac{d^2 y}{d x^2} = x^x \times \frac{1}{x} + \left( 1 + \log x \right)\frac{d}{dx}\left( x^x \right)\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = x^{x - 1} + x^x \left( 1 + \log x \right)^2 \left[ \text { Using } \left( 1 \right) \right] . . . \left( 2 \right)\]

\[ \therefore \frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x}\]

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

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

\[ = 0\]