Question

What is the derivative of (2x)^(2x)?

Answer 1

Ooo fun problem :o

Answer 2

We have a variable in the exponent, so we can't apply power rule.
We have a variable base, so we can't apply exponential rule.

What shall we do!! :)

Answer 3

Something to do with log..?

Answer 4

Yessss, something with log:)

Answer 5

Uhhh...... I would say take the log of both sides but there's nothing on the other side :/

Answer 6

We can `make up` another side if you want.
Say y=(2x)^(2x).

But there is another trick we can use.
It's really important to get comfortable with it if you're going to go on to higher math classes.

Answer 7

What's the trick?

Answer 8

So there is this idea that like...
\[\Large\rm e^{\ln(x)}=x\]If you take the composition of inverse functions, they "undo" one another.
So as I've posted above, this look familiar?

Answer 9

We want to use this idea in reverse.\[\Large\rm \color{orangered}{\left[(2x)^{2x}\right]}=e^{\ln\color{orangered}{\left[(2x)^{2x}\right]}}\]Or does this approach look too scary? :)

Answer 10

Don't the e and the ln cancel out because ln is the log of e?

Answer 11

Yes exactly, log base e, exponential function base e.
Cancel out isn't quite the right word.
But if you want to think of it in those terms, that's fine. :)

Answer 12

Okay, so then what do I do?

Answer 13

Don't they cancel out?*
They could, yes.
But we don't want to cancel them out.
We want to use that jumbled mess because it's easier to differentiate.

Answer 14

hmm \(\bf (2x)^{2x}\implies (2x)^x\cdot (2x)^x\)

Answer 15

\[\Large\rm \frac{d}{dx}e^{stuff}\]You remember how to differentiate exponential base e?

Answer 16

Uhh isn't it e^u(u')?

Answer 17

\[\Large\rm \frac{d}{dx}e^{stuff}=e^{stuff}(stuff)'\]Yes good.

Answer 18

\[\Large\rm \frac{d}{dx}e^{\ln\left[(2x)^{2x}\right]}=e^{\ln\left[(2x)^{2x}\right]}\frac{d}{dx}\ln\left[(2x)^{2x}\right]\]So we get something like that, yah?
Same exponential back.
Just need to multiply by derivative of the log.

Answer 19

The whole reason for doing this ... is that logs allows us to get variables out of the exponent position.

Remember your log rule? :o

Answer 20

Yes?

Answer 21

*still confused*

Answer 22

Too much? :c
Should we go back to the normal method you're used to?
Get lost in this method?

Answer 23

Kind of D:

Answer 24

\[\Large\rm y=(2x)^{2x}\]Taking the log of each side,\[\Large\rm \ln y=\ln (2x)^{2x}\]Ok applying our log rule,\[\Large\rm \ln y= 2x \ln(2x)\]From here we differentiate each side with respect to x.
Looks like we have to product rule on the right side.
Understand how to differentiate the left side?

Answer 25

(1/y) = (2x)(2/2x)+(2)(ln(2x))?

Answer 26

One thing to fix,
\[\Large\rm \frac{1}{y}\color{red}{y'}=2x\frac{2}{2x}+2\ln2x\]We're differentiating some variable with respect to x, so a y' should pop out, yes?

Answer 27

Yes, and then solve for y'?

Answer 28

yah :)

Answer 29

recall that \(\Large\rm y=(2x)^{2x}\)
So we'll need to plug that in for y at some point.

Answer 30

Got it!! Thanks so much :)