# ​If y = x^x, then prove that d^2y /dx^2 - 1/y(dy/dx)^2 - y/x = 0. ​

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If y = xx, then prove that $\frac{d^2y}{dx^2}-$$\frac{1}{y}$$(\frac{dy}{dx})^2$ - $\frac{y}{x}=0$

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Given,

y = xx

Taking logarithm on both sides, we get

log y = x. log x

Differentiating both sides, we get

Again differentiating both sides, we get