Derivatives of Composite Functions: The Chain Rule
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- Опубліковано 20 вер 2024
- Now we know how to take derivatives of polynomials, trig functions, as well as simple products and quotients thereof. But things get trickier than this! We may want to take the derivative of a composite function, where some function is operating on some other function. How can we do this? With the chain rule! It's easier than you think, I promise.
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Before watching : I know nothing about calculus
After watching : Wow, learning becomes easy when the tutor is amazing
We see. M
i guess I am pretty off topic but do anybody know a good website to watch new movies online?
@@rohansamuel7341 There is this thing called Google and if you type questions in it, guess what happens....you get answers, isn't that great...and correct too!
Thanks again Professor Dave! This was hurting my head a LOT from a text book but the colour sequencing and accompanying talkover is really helping me!
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True, knowledge is the holiest thing that one can receive from another 😉🙏 love from India
Professor Dave, you are an excellent teacher (and easy on the eyes, too). Thank you for your elephantine effort here on UA-cam...it's inspiring to a fellow (aspiring) polymath what your ambitions for this channel are. Bravo!
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You make me a great mathematician.
Wonderful! I'm starting a tutoring job later this month, so I'm reviewing all of calculus with your videos.
Wow, you made that look so simple. Best teacher ever!!! 👌🏾👌🏾👌🏾
Really well thought out and really well structured. Thanks as always.
Thank you so much for this! Your simplified explanation makes it easier for me to understand.
Before this I even didn't known the calculus but I watched all of your videos and know calculus very well as it is taught to my mind not to me now I am going to give JEE Mains and Adv and I think I am fully sure to solve these type of problems
how how your exams go? where are you in life now?
Got a different solution on the last comprehension exercise and thought I misunderstood and got it wrong, turns out I just didn't finish simplifying it (forgot to multiply the (x^2)(5x^4)) to get (5x^6). But I probably wouldn't even have gotten close without this video, so I really gotta say, great explanation!
Thank you for this comment I was having the Same Problem
If Dave was my teacher the only thing that would motivate me enough to come to school was just that he would be teaching and actually make the subject interesting
THANK YOU PROFESSOR DAVE OMG!!!!!
I got a math test on this and I was out the entire chain rule unit so YOU COMING IN CLUTCH
Thank you sir for your dedication and for making this free! 🙏
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in the last comprehension, prof dave i dont understand how u got 5x^6
Derivative of (Sinx^5) = cosx⁵+ derivate of x⁵= cosx⁵+ 5x⁴ now put that in equation
Same that’s the only thing that I’m confused about
Nvm I figured it out lol
The Chain Rule is about where my ability to learn calculus in high school and college petered out. Thanks to this tutorial (and a better understanding of recursion in general, plus playing with "chains" in Yugioh), I now fully understand differentiation. Thanks!
This is seriouslyyyy the best of the best , even you can't feel what I feel anymore, I am so happyyy omg lolllllllllll thanksssssss profffffff
Hey Professor, how did you get 5x^6 from x^2? (It's from Checking Comprehension Ex.3) I see you do the Product Rule first and then the Chain Rule but I can't see where it comes from.
so that's from when you take the derivative of sin x^5, chain rule gives you 5x^4(cos x^5) and then you combine 5x^4 with the x^2 you already have.
@@ProfessorDaveExplains Thank you very much for your quick response. It was a lot more simple than I thought!
Great explanation. The comprehensions are awesome.
how do you differentiate if the function is regular function and when it is a composite one? what is the primary characteristic to decide if the function is composite?
I took notes.
Thankfully there's a replay button.
Chain rule so easyyy! Thank you very much!
We need higher standards for college professors. We pay a lot of money for them to teach us while many don’t really teach us anything. The fact that we have to use google and UA-cam to learn more than they will ever teach us is ridiculous. I don’t know how, but this needs to change.
an amazing video, it has helped me immensely , just one question....where does the 5x^6 come from in the last step of the last question of check understanding ..im a bit confused
best and most easy explanation of chain rule on the internet!!!! thank you so much :)
Sir u r looking like one of Indian hero ranbir Kapoor 🙂and thank u so much for ur valuable lectures ty
i believe the answer of the 3rd question should have been 5x^4cos(sin(x^5)).(cosx^5)+2x.sinx(sin(x^5)) instead of 5x^6cos(sin(x^5)).(cosx^5)+2x.sinx(sin(x^5))
i believe you have made a typo
chances are you wont see this but what ever
Oh I got the same answer as you. Yeah, I think it's a typo.
I'm surprised that you leave the square root in the denominator at 3:34
Is it so easy he didn't bother showing or why don't we have to do it anymore?
To define "have to" depends on who's asking you to solve the problem. The reason why math teachers have you rationalize the denominator as a standard part of simplifying, is to make the problem more computationally friendly. If you didn't have a calculator, it is much easier to calculate sqrt(3)/3 instead of 1/sqrt(3) by hand, assuming you already know sqrt(3)'s digits.
It's also easier for a calculator or computer to handle calculations with rationalized denominators. A lot fewer steps, which can make your code run faster, if you've got a programming code that runs hundreds if not thousands of the same calculation.
In this case, because there is a variable under the square root, it is better to just leave it there. That way, you don't end up with more instances of the variable in your expression. If your goal were to solve for the variable, having fewer instances of it is an advantage.
This is the only explanation thats made sense
Sir please make videos on total derivatives
Well explained
May God bless you bro you are a real genius teach your help has become successful in my maths and chemistry grade
When using the chain rule, I usually prefer to start with the derivative of the inner function. Especially for multiple compositions, it often feels easier to start with the inner most function and work my way out.
Multiplication of derivatives is commutative.
Great explanation. Thanks Professor Dave.
I hope you all the best and I hope you good health . what is the derivative of y with respect to x for y= (x3+2x2+x)^3 ?
I suggest these steps to find dy by dx
1. bring down the power in the front of the bracket as a multiplier.
or bring down the power to the front of the bracket as a multiplier. it is better to put the preposition (in) or (to) in the above sentence .
2. reduce the power by one
3. multiply the derivative of the inside function.
dy/dx = 3 (x3+2x2+x)^2 (3x2+4x+1)
Would you please recommend and advise me what are the best method to find dy by dx ?
All the best
Moamen Saeed Talha
Could explain need of derivative and formation of trigonometric function's derivative a bit more in detail ?
that's in a previous clip! check my math playlist or calculus playlist, should be called something like derivatives of trigonometric functions
Will you please show the graphical representation for chain rule and product rule
Using graphs isn't really the best way to visualize those two. If you want, you can watch 3B1B's video about that: ua-cam.com/video/YG15m2VwSjA/v-deo.html It's also an excellent video, especially if you're a visual learner.
I had a lot of trouble understanding the 3rd comprehension question. Even with the answer, I couldn't work backwards through that to see how I was going wrong. I had to find a differentiation calculator which showed how the differentiation rules were applied in a step by step manner to understand how I was going wrong.
omg, thanks!! you are the best ever; ❤❤❤
Great video!! From grade 4 and grade 6 students from Canada
How did you get that 3cosX in the second example (comprehension).
Please reply
that term in brackets is the product rule, the second term is the derivative of 3x, which is 3, times the other term, cos x.
@@ProfessorDaveExplains thank you
in the comprehension number 2.) can I factor the coefficient 3 from (-3xsinX + 3cosX) and put it on the left side?
Excellent work.
in the second example wasnt the first one sinx so if the derivative of sin of x is cosx why is it cosx(squared) ??? can someone explain
because it's sin(x^2) so we do the outer function first, sin becomes cos, and then we multiply by the derivative of the inside.
3rd homework question is a good one
Please provide video on complex numbers sir
i did that already!
Thank you
Oh my god HTANK YOU SO MUCH
You’re very good
6:06 would anyone explain why he combines those two terms, they don't seem related to me.
In the last problem in Checking comprehension why
(x^2) [cos(sin x^5) . (sin x^5)' ] turned into
5x^6 cos(sinx^5) . (cos x^5) I mean why the x^2 turned into 5x^6
maybe i'm late but that happens when u go for derivative of sin x^5, where you apply chain rule to get cos x^5 (x^5)'
and that derivative of x^5 is 5x^4 which you multiply with that x^2 outside to get that 5x^6
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whats the name of this playlist for all the vids , plz
thank u i love u
these are all in my calculus playlist, as well as the longer mathematics playlist
hey can you upload gauss law please
d/dx of sin²(x) = sin (2x) 🎉
Where does the 5x^6 came from ?
It comes from your UHsSS
Professor Dave, the volume of your videos are relatively too soft. I have noisy ass neighbours and I can only put out that much at maximum with this little pc i got. no earphones at all
Done.
5:40
I'm stuck at finding the derivative of a number to the x
derivatives of exponential functions coming up on monday!
In a situation where there is an exponential function in the form g(x)= a^f(x), In which a is a real number, the formula for the derivative is as follows.
g'(x)=a^f(x)*lna*f'(x).
So you first write the original exponential function down, then you multiply by the ln of the base, and lastly, multiply by the derivative of the exponent. If the exponent was simply x, you would multiply by 1 since that is the derivative of x. If the exponent was 2x, you would multiply by 2. If it was sin x, you would multiply by cos x and so on.
@@reubenmanzo2054 You can differentiate i^x. The OP just limited the statement to real numbers, to keep it simple, as a lot more can happen with complex numbers that you might not anticipate. Therefore, you generally avoid acknowledging that a rule also works for complex numbers if that is beyond the scope of a particular class or topic. You then limit your domain to real numbers so that you don't accidentally give an incorrect statement for how the rule works with complex numbers.
To differentiate i^x relative to x, you rewrite it in its alternative form of e^((i*pi*x)/2). Define k=i*pi/2, and rewrite as e^(k*x). Now differentiate with the chain rule, and get k*e^(k*x). Recall the value of k, and rewrite. You will get: 1/2*i*pi*e^(1/2*i*pi*x), which you can replace the base e term with the original expression i^x, and write as 1/2*i*pi*i^x.
@@reubenmanzo2054 My method shows that this works if you didn't already have established how to determine ln(i). The k introduction was optional, and I just chose to do that so the equations could be compact in the intermediate steps.
W professor
Excellent explaining..but my dumb brain only got to understand the musical intro..🤣
my head hurts
I am broke but you still deserve my money
Hii
Don't put chains on me.
The subtitles won't let me see it
so turn them off
@@ProfessorDaveExplains Okay, thanks.
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Jesus Thank you
it’s fine as long as the exponent belongs to R
The power rule works for complex exponents as well. They just don't bother teaching this to you in introductory Calculus, to keep it simple.
Sir please I don't still understand chain rule
Thanks a lot for the tutors I find it very helpful.