What is the derivative of $(2x)^{4x}$? This is quite simple. I know. I am having a problem when comparing my answer to online calculators like Symbolab and such. 
$$\begin{array}{rll} y &= **(2x)^{4x}** & \text{equation}\\
\ln(y) &= \ln((2x)^{4x}) &\text{ take ln of both sides to bring 4x out front}\\
\ln(y) &= 4x  \ln(2x) & \text{ use log property}\\
1/y * y' &= 4\ln(2x) + 4x  (1/2x)  2 & \text{ use product and chain rule}\\
y' &= y ( 4\ln(2x) + 4x (1/2x)  2 ) &\text{ multiply both sides by y}\\
y' &= 2x^{4x}( 4\ln(2x) + 4) & \text{simplify }4x*2 = 8x / 2x = 4 \\
y' &= 8x^{4x}\ln(2x) + 8x^{4x} &\text{further simplify}\\
y' &= 8x^{4x}(\ln(2x) + 1).&
\end{array}$$
However, the answer on symbolab gives $8x^{4x}(\ln(x) + 1)$. <--- Disregard that. That was based on my incorrect input of the first line. 
Am I wrong? If so, how? 

Wrong on the third line, $\ln(2x^{4x})=\ln2+4x\ln x$ but you confused with $\ln((2x)^{4x})=4x\ln 2x$.
