# What is the derivative of sqrt(x ln(x^4))?

$\sqrt{x \ln \left({x}^{4}\right)} = {\left(x \ln \left({x}^{4}\right)\right)}^{\frac{1}{2}}$, so the derivative is $\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x \ln \left({x}^{4}\right)}}$.
So the derivative is, $\left(1\right) \ln \left({x}^{4}\right) + \left(x\right) \left(\frac{1}{x} ^ 4\right) \left(4 {x}^{3}\right) = \ln \left({x}^{4}\right) + 4 {x}^{4} / {x}^{4} = \ln \left({x}^{4}\right) + 4$.
1/(2sqrt(xln(x^4)))(ln(x^4)+4)=(ln(x^4)+4)/(2sqrt(xln(x^4)).