Given, y = x^x ⇒log y = x log x ⇒1/ydy/dx=1+log x ⇒dydx=x^x(1+log x) ∴ y'(1) = 1(1+log 1)=1 ...

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Logarithmic Differentiation

Find y=xx, ...

Question

Find $y = x^{x}$ , find $y^{'} (1)$

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Solution

Given, $y = x^{x}$
$\Rightarrow log y = x log x$
$\Rightarrow \frac{1}{y} \frac{d y}{d x} = 1 + log x$
$\Rightarrow \frac{d y}{d x} = x^{x} (1 + log x)$
$∴ y^{'} (1) = 1 (1 + log 1) = 1$

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