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Product rule proof | Taking derivatives | Differential Calculus | Khan Academy
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- เผยแพร่เมื่อ 4 มี.ค. 2015
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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.
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What can I say....brilliant explanation of the product rule for derivatives.
hi
@@zadocktoo3652 what?
thank you - much better explanation than what i was looking at in my calculus course material.
I knew that the only one who could explain this is the blessed Mr. Khan.
without you, I wouldn’t have such knowledge in math and calculus
تخرجت ؟
@@Mohd-l اي :)
@@MohaMMaDiN55 تشتغل حاليا؟
Exceptional Mr Khan. You are appreciated in South Africa!
waarheid
Second that 🙏
One of the sexier proofs, IMO. :D
Definitely elegant
I never fully understood this during my A-Levels 10 years ago, and had to just memorise the final formula. Now I get it!! Thank you so much :D
This also proves integration by parts just by doing a little bit more algebra
i am gonna be a tutor like you Mr. khan . you are my inspiration
Great!!!!!!!
For those of you that want to practice some exercises, in my latest video I go through exactly these exercises step-by-step, explaining everything I do along the way. Perhaps it helps some of you :)
"If at any point you feel inspired, I encourage you to pause this video."
Spoken while my finger was on the space bar.
Great time to try and prove as the only thing that is not intuitive is that one first algebraic manipulation, which would seem totally arbitrary if we didn't know it was necessary to prove the product rule.
I really appreciate your effort mr khan
Logarithmic differentiation is far more efficient here.
I really think the chain rule should be taught much earlier. With just the sum and constant rules, the derivative of sine, and the derivative of ln, one can easily differentiate any function.
Whenever I teach this to anyone I will write at the end, - by Sal Khan
Someone get this guy a drummer.
Why audio is so low?
why they never teach this in any calculus course?? like it's fundamental
Thank you so much, my Professor skipped so many steps😭
Excellent as always! U are a great teacher! May I humbly suggest: A transcript of the blackboard notes could be very useful.
We copy out Ur blackboard longhand anyhow. Much simpler to just copy/paste a transcript...
I imagine Ur digital blackboard already generates them..
Maybe it's already available but I just dont know it?
Where did you get f(x+h)g(x) ? I get it cancels out, but where did that come from?
I really hope it was trial and error as I am unsettled by the thought that somebody could intuit that applying +f(x+h)g(x) -f(x+h)g(x) would allow you to derive the formula...
Firstly,... Brilliant explanation of the rule,helped me a lot but I still have a doubt about taking f(x+h) as f(x)
Is there a way to generalize the product rule, as in a rule considering the derivative of the product of three functions or more instead of just two?
Yes just apply the product rule twice
Thanks a lot !
Easy way to remember product rule: Keep diff + diff keep
That was a very nice approach I learnt. Thank you.
Easier if f(x+h) depicted as f(x)+df and g(x+h) as g(x)+dg
Then THEIR product is simply: f(x).g(x) +f(x).dg +g(x).df +df.dg
Then subtracting f(x).g(x) and dividing by dx gives:
f(x).dg/dx +g(x).df/dx +df.dg/dx
or f(x).g' + g(x).f' while the remaining term df.dg/dx is removed as it contains ALL the averaging error (not because it is zero)
BTW, since f(x+h) is f(x) + df
So f(x+h). g' = f(x).g' + df.g'
Then df.g' is eliminated as it contains averaging error and NOT because it is zero.
Thank you Sal for your nice explanation, my daughter loves your explanation 🙏🙏🙏
Thank you!!! I love proofs like these so I don't just need to blindly trust that it works! :) Great content!
wow, u have saved my a lot of times
great thx!
What gives you the "right" to factor out f (x+h) from the numerator? I understand why you can factor out g(x), since it has no h in it. But f(x+h) has h as a numerator. How do we justify removing the denominator from it?
6:25 he read my mind
So much easier than Wikipedia
Smart!
Beautiful! Thank you.
Thx from india... Just another jee aspirant
amazing explanation. thank you.
this is a brilliant explanation! thank you
Excellent 👍
Thnx for nice explaination...
wow! Masha allah , thank you mr.Khan
You are a lifesaver!!!!
Very well explained, well done
Very clever.
Well proven
this is magic
Great
Thanx bro
Lol.. I get that what he mean now. Excellent explaination.
Thank you for the video but I think this proof is not correct because
lim as h->0 of f(x+h) is Not equal to f(x). That is the same with saying that lim as x->a of f(x) = f(a) which is not true. That being the whole point of the limit, to show that the value that your are approaching is not the same with the value of the function at that point.
Think of it like this: looking for the value that f(x+h) approaches when h->0 is the same with looking for the value that f(n) approaches when n->a (some random value a). The values that f(n) takes when is goes closer and closer to a are not neccesarly closer and closer to f(a). That's why there is a value for f(a) and another value for the limit.
I hope it's not pure nonsense what I ve just wrote.
Fantastic ❤️✨
perfect
our Saviour!
Audio ?????
This is one of the proof. So is there another proof that is also valid. I'm just curious
Leibniz has one. Was so excited he wrote his mom about it. I like it and it's graphical using rectangles
please, sent any link or source we can find that
Who came here from MIT lecture? You know what I mean :'D
@Rev limits shut up, it's just a joke
@Rev limits ?
@Random Autonomous Drone Pilot Hi khan academy is great ofc we'd watch it .
Also dont forget life savers like Sal in here is of Bangladeshi descent , so if you want pure american content go pay the fortune to get into harvard or something
@Random Autonomous Drone Pilot Good for you :)
and who exactly are the "we" you're assuming me to be ?
it’s not a very satisfying proof imo
I don't think I like the way the new "pen" writes