Question

y = x^{2} + (logx)ˣ


By using logarithm differentiation.

Answer 1

$u = {x}^{2} \\ \frac{du}{dx} = 2x$
$v = { log(x) }^{x} \\ log(v) = x \: log( log(x) )$
differentiate w. r.to x

$\frac{1}{v} \frac{dv}{dx} = x \: \frac{1}{ log(x) } \frac{1}{x} + log( log(x) )$
$\frac{dv}{dx} = { log(x) }^{x} ( \frac{1}{ log(x) } + log( log(x) )$

therefore

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$