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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?

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  • $\begingroup$ The answer in symbolab is different. I guess your formatting was wrong even there. $\endgroup$ – Manish Kundu Feb 7 '18 at 7:44
  • $\begingroup$ I am aware. I have already messed this post up quite a bit. I updated the first line but did not change the result of what symbolab gave. @ManishKundu $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 7:50
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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. :)

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  • $\begingroup$ I edited the problem I posted. It was supposed to be (2x)^(4x), excuse my untidy formatting. $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 7:31
  • $\begingroup$ @MichaelRamageMikeRamage edited answer. now check :) $\endgroup$ – Manish Kundu Feb 7 '18 at 7:44
  • $\begingroup$ Now I am confused as to where the 2^(4x+2) came from, specifically the +2. Is there a rule I have totally forgotten about? @manishkundu $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 7:49
  • $\begingroup$ $(2x)^{4x} = 2^{4x} . x^{4x} $ $\endgroup$ – Manish Kundu Feb 7 '18 at 7:56
  • $\begingroup$ Sorry that ln x was a typo $\endgroup$ – Manish Kundu Feb 7 '18 at 7:57
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Wrong on the third line, $\ln(2x^{4x})=\ln2+4x\ln x$ but you confused with $\ln((2x)^{4x})=4x\ln 2x$.

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  • $\begingroup$ I updated the problem. I was not aware that I posted it like that. @Sujit Bhattacharyya $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 7:34
  • $\begingroup$ it's ok. don't need to worry $\endgroup$ – Sujit Bhattacharyya Feb 7 '18 at 8:05
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$(2x)^{4x}=e^{4x\ln(2x)}$, 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$[(2x)^{4x}]'=[e^{4x\ln2x}][4\ln(2x)+\frac{4x}{x}]=(2x)^{4x}[4\ln(2x)+4]$

$${\color{red}{(2x)^{4x}\neq 2x^{4x}}}$$

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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: $$y’=((2x)^{4x})’=\left((2x)^{4x}\left(4\log(2x)+4x\frac{2}{2x}\right)\right)=\\=4\cdot(2x)^{4x}(\log(2x)+1)=\color{red}{4^{2x+1}x^{4x}(\log(2x)+1)}$$

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  • $\begingroup$ The power 4x belongs to both the 2 and x (2x)^(4x). I had to make the edit. I did not realize my mistake. @Pippo $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 7:40
  • $\begingroup$ I edited my answer @MichaelRamageMikeRamage $\endgroup$ – user507623 Feb 7 '18 at 7:57
  • $\begingroup$ I appreciate the help Pippo. I thank you and all those who put up with my errors in input. $\endgroup$ – Michael Ramage MikeRamage Feb 7 '18 at 8:09
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$y=(2x)^{4x}$

$y=e^{\ln (2x)^{4x}}=e^{4x\ln (2x)} \implies \frac {dy}{dx}=(4\ln (2x)+4x\cdot \frac{2}{2x})\cdot e^{4x\ln (2x)})=4(\ln (2x)+1)(2x)^{4x}$

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  • $\begingroup$ Why downvote the simplest and direct way of solving this? There was just a typo in the beginning. I'm upvoting this. $\endgroup$ – Ken Draco Feb 10 '18 at 23:18
  • $\begingroup$ And the typo was on the part of the OP in the first place! See the edits, please. $\endgroup$ – Ken Draco Feb 10 '18 at 23:25

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