#Derivative of aˣ for any positive base a

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nimble cove
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I'm having trouble understanding the video containing the following comment and if it's correct then I feel I could grasp the video using that comment. Could you confirm the comment is correct and if you could add something to make it clearer it would be great. Notice you will have to scroll up a bit to see the beginning of the comment, by Michael Hopkins. https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-1b/v/exponential-functions-differentiation-intro?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfODAwNTk3NzQ5NjE5MTU1NTg1MTg1ODkwDAsSCEZlZWRiYWNrGICAgOCTt4oKDA&qa_expand_type=answer
If you think the comment is incorrect and you could clarify the video, notice the tab transcript, it would be appreciated.

fading bearBOT
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unreal kestrel
nimble cove
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My question is directly related to that video.

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again, that the transcript on that page

unreal kestrel
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Just tell me what was said in the video that has you so confused.

nimble cove
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From 2:56 I don't understand, the transcript is timestamped.

unreal kestrel
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Then copy/paste it.

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We have literally spent more time arguing about what's in the video than it would have taken you to just say what's in the video.

nimble cove
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2:56And now we can use the chain rule
2:59to evaluate this derivative.
3:02So what we will do is
3:05we will first take the derivative of the outside function.
3:10So e to the natural log of a times x
3:14with respect to the inside function,
3:17with respect to natural log of a times x.
3:21And so, this is going to be equal to e
3:24to the natural log of a times x.
3:28And then we take the derivative of that inside function
3:31with respect to x.
3:33Well natural log of a,
3:34it might not immediately jump out to you,
3:36but that's just going to be a number.
3:38So that's just going to be,
3:39so times the derivative.
3:41If it was the derivative of three x, it would just be three.
3:44If it's the derivative of natural log a times x,
3:46it's just going to be natural log of a.
3:49And so this is going to give us
3:52the natural log of a times e
3:56to the natural log of a.
3:59And I'm going to write it like this.

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4:01Natural log of a to the x power.
4:04Well we've already seen this.
4:08This right over here is just a.
4:11So it all simplifies.
4:13It all simplifies to the natural log of a
4:17times a to the x,
4:21which is a pretty neat result.
4:23So if you're taking the derivative of e to the x,
4:25it's just going to be e to the x.
4:27If you're taking the derivative of a to the x,
4:30it's just going to be the natural log of a times a to the x.
4:34And so we can now use this result
4:36to actually take the derivatives of these types
4:40of expressions with bases other than e.
4:43So if I want to find the derivative
4:46with respect to x
4:48of eight times three to the x power,
4:52well what's that going to be?
4:54Well that's just going to be eight times
4:57and then the derivative of this right over here
5:00is going to be, based on what we just saw,
5:03it's going to be the natural log of our base,
5:05natural log of three times three to the x.
5:09Times three to the x.
5:12So it's equal to eight natural log
5:15of three times three to the x.
5:19Times three to the x power.

unreal kestrel
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Do you not know what the chain rule is?

nimble cove
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u(v(x)) = u(v(x))*v(x)

unreal kestrel
nimble cove
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I can't use the prime character!

unreal kestrel
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Yes you can, because it's an apostrophe, not a backtick.

nimble cove
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u(v(x)) = u'(v(x))*v'(x)

vestal viper
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techie it’s fine

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he’s doing the best he can

unreal kestrel
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u(v(x))' = u'(v(x)) * v'(x)

unreal kestrel
nimble cove
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If you don't help with the quote, I'll just assume the comment is correct, with the 27 upvotes it have.

vestal viper
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your method ig

vestal viper
unreal kestrel
nimble cove
vestal viper
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dude

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can you just post the damn comment instead of linking the video

unreal kestrel
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Screenshot.

vestal viper
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i can’t open my browser rn

nimble cove
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Michael Hopkins
6 years ago
Posted 6 years ago. Direct link to Michael Hopkins's post “Ok, after going over this...”
Ok, after going over this again and again, I believe I have worked this out.

Let:
f(u(x)) = e^u(x)
u(x) =ln(a)*x
G(x) = e^(ln(a)*x) = f(u(x))

f'(u) = e^u (using the derivative of e rule)
u'(x) = ln(a) (using constant multiple rule since ln(a) is a constant)

so G'(x) = f'(u(x))*u'(x) (using the chain rule)

substitute f'(u) and u'(x) as worked out above
G'(x) = (e^u(x))*ln(a)

substitute back in u(x)
G'(x) = (e^(ln(a)*x))*ln(a)

towards the beginning of the video, Sal determined that a = e^ln(a), so this can be substituted into the above equation of the the final answer of:

G'(x) = (a^x)*ln(a)

Hopefully they way I've written in out helps and doesn't cause any confusion!
Comment on Michael Hopkins's post “Ok, after going over this...”
(27 votes)

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vestal viper
nimble cove
vestal viper
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if you want me to demonstrate it more visually i can

unreal kestrel
nimble cove
vestal viper
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$a^x=e^{xln(a)}$

unreal kestrel
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No it doesn't.

vestal viper
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...

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my bad

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typo

analog prismBOT
nimble cove
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yes, it's on the video

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at 2:56

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I understand up to 2:56

vestal viper
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let’s just call the derivative of a^x dy/dx for simplicity

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then using chain rule

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where $(f(g(x)))’=f’g(x)*g’(x)$

nimble cove
analog prismBOT
vestal viper
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dude

unreal kestrel
vestal viper
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don’t say that

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he’s teaching you very slowly

nimble cove
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DynV
OP
— Today at 4:16 PM
yes you can see it at 2:56, which I mentioned I understand up to there

vestal viper
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let me continue

unreal kestrel
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Why are you so anti-communication?

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Actually, I don't care, I'm out.

vestal viper
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$f(x)=e^x$

analog prismBOT
vestal viper
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$g(x)=x(ln(a))$

analog prismBOT
vestal viper
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now $f’(g(x))=e^{x(ln(a))}$

analog prismBOT
vestal viper
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and $g’(x)=ln(a)$

analog prismBOT
vestal viper
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thus $(f’(g(x)))’=e^{x(ln(a))}*ln(a)=(ln(a))a^x$

analog prismBOT
vestal viper
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@nimble cove does this make sense

nimble cove
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I don'T understand g(x) to g'(x)

vestal viper
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wdy

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m

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its literally the chain rule

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$(f(g(x)))'=f'(g(x))*g'(x)$

analog prismBOT
vestal viper
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and if youre talking about x(ln(a)) to ln(a)

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its because of the constant multiple rule

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where $\frac{d}{dx} (cx)=c$

analog prismBOT
nimble cove
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I'm sorry, perhaps I'm having a brain fart or too tired but I don't understand g(x)=x(ln(a)) to g’(x)=ln(a).

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I read what you wrote from 5 mins ago,.

nimble cove
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Well. Thanks for confirming the comment was correct.

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+close 5

fading bearBOT
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