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
I know it's a year late but that uses Implicit differentiation. I know it's a special case of chain-rule but if this video is for say High-Schoolers, they might have not learned that case yet.
@@jacksonzheng3103 do you mean in the final answer? If so, you would sub x back in so that you are using the variables given in your problem. y is just a strategy.
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!
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
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.
REIN Cenollari I think the exact same...:( lol it looks soooooo complicated laid out like this. But still, thanks for the help. :) I might have simplified a few things and included that after using the chain rule to account for the first part times the derivative of the inside (inside being the ln 2 multiplied by x) you use the product rule for finding the derivative of the ln 2 multiplied by x.... which then gives you the ln (2) at the end that is multiplied by 2^x. But still thanks again!..:)
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)
What he's doing is the proper way to do it; showing you exactly why the proof is the way it is. The way he does it is the proper mentality when doing maths, systematically breaking it down into easier formulae, so that we can solve them.
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)
@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
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.
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?
Can you please tell me the software and device use to write digitally.....i am a teacher and have back pain problem so i am searching something that i can sit and write and project it on board
The software is SmoothDraw3. Sal writes on a Wacom Bamboo Tablet. Sorry that you didn't get a response in four years. I wish you the best in your teaching endeavors.
Kuch samaj nhi aaya.. please speak without any style.. this just shows a style in ur voice nothing else.. Neither I understood anything nor I liked it..
Well... not everyone learns at the same pace as you as i understood this 100% For those who work at slower paces, they should always ask questions rather than saying "you suck" cause teachers arent mind readers...
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
I know it's a year late but that uses Implicit differentiation. I know it's a special case of chain-rule but if this video is for say High-Schoolers, they might have not learned that case yet.
@@TheDannytaz I know it's two years late but this chain rule is known to us highschoolers:)
why would you write x in place of y? could anyone explain?
@@jacksonzheng3103 do you mean in the final answer? If so, you would sub x back in so that you are using the variables given in your problem. y is just a strategy.
Knowing implicit that makes more sense than the video. Thanks.
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.
I love your proof. I wonder how did you come up with it? I never thought of it!
QED is quantum electrodynamics
@@josephlau13d77 not exactly
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!
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
That was what I was thinking
thank you so much for this ! that makes a lot more sense to me
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
Don't understand people complaining, he made this stuff dumb easy now.
For me, this tells me what I need, to understand the whole idea about it.
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.
shut up
your confusing me even more
stera Perowski
Me too.
same here
Ur just dumb
Watch again and listen more carefully
i love the comment section, always have the first thing in mind LOL
omg you make it so complicated when it is so easy -_-
Trust me, all that notation is really really important down the road
REIN Cenollari I think the exact same...:( lol it looks soooooo complicated laid out like this. But still, thanks for the help. :) I might have simplified a few things and included that after using the chain rule to account for the first part times the derivative of the inside (inside being the ln 2 multiplied by x) you use the product rule for finding the derivative of the ln 2 multiplied by x.... which then gives you the ln (2) at the end that is multiplied by 2^x. But still thanks again!..:)
MistahJuicyBoy what are you talking about down the road? Integrals?? I mean I’m already in calculus what’s next???
@@jmshkm Calc 2
@@jmshkm Depends on your focus. If it's engineering, you've at least got differential equations and linear algebra next.
I really appreciate this. I know the rule but I like to see the math that makes it true. Great work thank you
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)
What he's doing is the proper way to do it; showing you exactly why the proof is the way it is. The way he does it is the proper mentality when doing maths, systematically breaking it down into easier formulae, so that we can solve them.
What a great explanation. You make it look so simple.
Awesome. Thanks for the help. If it wasn't for Khan Academy I'd have to sink countless hours into the Calculus book.
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)
this ain't the question bro. It's 2^x not 2/x
@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
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.
Like Mohammed said:
(a^x)' = a^x * ln a * x'
The " x' " is just the chain rule! Understanding why it's the case is useful though, so this vid is great
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?
Why u no implicit differentiation?
@Abdo Kaboobie hahha
just skip to the end if you don't want the proof
(a^f(x))' = a^f(x) * (f(x))' * ln(a)
Can you please tell me the software and device use to write digitally.....i am a teacher and have back pain problem so i am searching something that i can sit and write and project it on board
The software is SmoothDraw3. Sal writes on a Wacom Bamboo Tablet.
Sorry that you didn't get a response in four years. I wish you the best in your teaching endeavors.
just take the natural log from both sides..
Well explained. Thanks a lot!
Mr. Khan I love your vids please do one on science experiments please? Thank you and please reply :)
Thanks for the help and thank you Wisdom for sharing this to me :)
Thanks for the help.
Thanks bro
You make it so easy 😎
Bangali?
I got it after long time😁
Just compare it with a^x
khan academy ,i just want to know to produce electricity from chlorophyll.my friend did it so csn you tell how is it possible?
Thank you!
thanks
I love ur videos!:)
what about the derivative of 2/x ???
Krystle Lopez x X
Krystle Lopez xxx
-2/x^2
Nice
what if the x is squared?
treat it as 2^(x*2)
Love it :D
Plz don't use pink colour
Way too confusing
It's cool 😎 😎 😎 😎
Tf???
👍👍👍👍👏👏👏👏👏❤️❤️❤️💐💐💐
First to comment too!
absolute terrible explaination
Kuch samaj nhi aaya.. please speak without any style.. this just shows a style in ur voice nothing else.. Neither I understood anything nor I liked it..
this guy is so annoying, not helpful
What's with all these tutorials and writing all this crap with "e" when they can just derive implicitly ?
Waist channel
Khans academy sucks at explaining -_-
Well... not everyone learns at the same pace as you as i understood this 100%
For those who work at slower paces, they should always ask questions rather than saying "you suck" cause teachers arent mind readers...
@@winterwraith8339 you are awesome
thank you!
thanks