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? 
 A: 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. :)
A: Wrong on the third line, $\ln(2x^{4x})=\ln2+4x\ln x$ but you confused with $\ln((2x)^{4x})=4x\ln 2x$.
A: $y=(2x)^{4x}$
$y=e^{\ln (2x)^{4x}}=e^{4x\ln (2x)} \implies \frac {dy}{dx}=(4\ln (2x)+4x\cdot \frac{2}{2x})\cdot  e^{4x\ln (2x)})=4(\ln (2x)+1)(2x)^{4x}$
A: You have to remember that the derivate of a power can be expressed as
$$f(x)^{g(x)}=e^{g(x)\log f(x)}\\(f(x)^{g(x)})’=f(x)^{g(x)}(g(x)\log f(x))’=f(x)^{g(x)}\left(g’(x)\log f(x)+g(x)\frac{f’(x)}{f(x)}\right)$$
Then you have simply to do this:
$$y’=((2x)^{4x})’=\left((2x)^{4x}\left(4\log(2x)+4x\frac{2}{2x}\right)\right)=\\=4\cdot(2x)^{4x}(\log(2x)+1)=\color{red}{4^{2x+1}x^{4x}(\log(2x)+1)}$$
A: $(2x)^{4x}=e^{4x\ln(2x)}$, so $\frac{{\rm{d}}((2x)^{4x})}{{\rm{d}x}}=\frac{{\rm{d}}(e^{4x\ln(2x)})}{{\rm{d}x}}$.
now,we have $[e^{f(x)}]'=e^{f(x)}f'(x)$.
$[4x\ln(2x)]'=4\ln(2x)+\frac{4x}{x}$
so$[(2x)^{4x}]'=[e^{4x\ln2x}][4\ln(2x)+\frac{4x}{x}]=(2x)^{4x}[4\ln(2x)+4]$
$${\color{red}{(2x)^{4x}\neq 2x^{4x}}}$$
