Answer
$y^\prime=(2x)^{2x}(2ln(2x)+2)$
Work Step by Step
If we let the function we are trying to differentiate be equal to y we have: $y=(2x)^{2x}$, now we can use logarithmic differentiation. Start by taking the natural log of both sides: $ln(y)=ln((2x)^{2x})=2xln(2x)$ Now if we differentiate both sides using the product rule we get: $\dfrac{y^\prime}{y}=2ln(2x)+2$ now we need to multiply both sides of the equation by y which is just $(2x)^{2x}$ so: $y^\prime=(2x)^{2x}(2ln(2x)+2)$
