# How do you differentiate f(x)=x^(x^2)?

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Sep 14, 2016

Let $y = {x}^{{x}^{2}}$

Taking log on both sides,

$\log y = \log {x}^{{x}^{2}}$

By logarithmic properties,

$\log y = {x}^{2} \cdot \log x$

Differentiating the equation using chain rule and product rule,

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \cdot \log x + {x}^{2} / x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{{x}^{2}} \left(2 x \log x + x\right) = {x}^{{x}^{2} + 1} \cdot \left(\log {x}^{2} + 1\right)$

If you have any kind of confusion or want me to elaborate in the steps of solving, comment below.

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