How do you find the derivatives of $y=x^(2x)$ by logarithmic differentiation?

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Answer 1 · 2017-02-11T15:34:19

The answer is $=2(1+lnx)x^(2x)$

We need
$(uv)'=u'v+uv'$

$y=x^(2x)$

$lny=ln(x^(2x))$

$lny=2xlnx$

Differentiating wrt $x$

$1/ydy/dx=2(x*1/x+1*lnx)$

$dy/dx=2(1+lnx)y$

$dy/dx=2(1+lnx)x^(2x)$

How do you find the derivatives of y=x^(2x)y=x^(2x) by logarithmic differentiation?