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? 

Given, 
$$ f(x) = y =  (2x)^{4x}$$
Taking natural log on both sides, 
$$ \ln y = 4x \ln 2x$$
Now differentiating w.r.t. x,
$$ \frac{1}{y}\frac{dy}{dx} = 4x.\frac{1}{2x}.2 + ln 2x . 4$$
$$ \frac{dy}{dx} = y \,(4+4\ln 2x)$$
$$ = 4(2x)^{4x}(1+\ln 2x)$$
$$ = 2^{(4x+2)} . x^{4x} (1+ \ln 2x) $$
That's even the answer given in symbolab. :)
