Chain rule for derivative of 2^x

[aryanarora]

there is a simpler way, as follows:

let y = 2^x

apply natural log on y

=> ln y = x. ln 2

differentiate ln y w.r.t "x"

=> 1/y . dy/dx = (1).ln 2 + x(0)
=> 1/y . dy/dx = ln 2
=> dy/dx = y.ln 2

plug y=2^x in dy/dx

=> dy/dx = 2^x.ln 2

[animimm]

let y=2^x implies lny=ln(2^x)=xln2 now differentiate both sides with respect to x:
y'/y = ln2 implies y'= yln2= (2^x)ln2 Q.E.D.

[forgotaboutbre]

Perhaps my explain may be a bit simpler:


Once you have y = [e^(ln2) ] ^ x, apply the derivative with rules for base e.


dy/dx =ln2* [e^(ln2)] ^ x
= ln2 * [ 2 ] ^ x, where e^(ln2) was replaced by 2


Tidy it up and


df/dx = ln2 * 2^x

[anoniem]

For me, this tells me what I need, to understand the whole idea about it.

[venus]

I really appreciate this. I know the rule but I like to see the math that makes it true. Great work thank you

[FishyBusiness]

Everyone in the comments afraid of something called a proof. Yea there's a rule around deriving these but he's just showing where that rule comes from

[platinumtrio]

you know if you wanted to just memorize the simple equation for this you shouldn’t have clicked on a 5 minute video... i can never memorise it so this was super helpful!

[rickhoro]

What a great explanation. You make it look so simple.

[mohammedabid]

guys the rule is (a^x)' = a^x * x' * ln(a), so in this case it would be ( 2^x)'= 2^x * 1 * ln(2). so Y'= (2^x)(ln(2)). This video is making it hard for no reason it just confused the shit out of me.

[nebvbn]

General rule for when (bx + c) is the exponent:
Let f(x) = a^(bx + c)
f'(x) = d/dx(a^(bx + c))
= d/dx(e^((bx + c)ln(a)))
= e^((bx + c)ln(a)) * d/dx(bxln(a) + cln(a))
= e^((bx + c)ln(a)) * bln(a)
= a^(bx + c) * bln(a) <--(answer)
Therefore all you have to do is to multiply the original function with the coefficient of x and the natural log of the base of the exponent.
An even more general rule:
Let f(x) = a^g(x)
f'(x) = a^g(x) * g'(x)ln(a)

[renaldocenollari]

omg you make it so complicated when it is so easy -_-

[igotbev]

Don't understand people complaining, he made this stuff dumb easy now.

[coffeetablezdj]

@Nathan Guhl if x is squared then 2/x^2 is basically 2*x^-2 multiply the statement by the exponent and lower by 1....you should get -4*x^(-3)....
umm...you are welcome

[fabse]

If you think about it, 2/x is really just another way of writing 2*x^(-1). Now you just have to multiply the statement by the exponent and lower it by 1, and you should get: -2*x^(-2)

[enkidu_]

Awesome. Thanks for the help. If it wasn't for Khan Academy I'd have to sink countless hours into the Calculus book.

[rockfordlines]

The explanation was pretty good, but it would have been better to spend 20 seconds more on the derivation of the chain rule so that everyone would understand a bit better.

[wediadi]

So this can be generalized like if you want to calculate the derivative of a costant raised by 'x', then the result will be the same function multiplied by the natural logarithm of the constant?

[colinkiama]

Thanks for the help and thank you Wisdom for sharing this to me :)

[breisfm]

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

[lameesazahedul]

Mr. Khan I love your vids please do one on science experiments please? Thank you and please reply :)