This video will show you how to do the product rule for derivatives. Remember to use this rule when you want to take the derivative of two functions being multiplied by one another. For more videos please visit www.mysecretmat...
This program (and the associated comments below) clearly demonstrate that most people have problems with math due to bad teaching. Here the presentation is clean, orderly, divided into just the right "chunks" and, to top it off, delivered in a pleasing voice. Great job!
How can I spend 3-4 hours in lectures and be clueless, then turn around and spend 1 hour on this channel and know everything I didn't understand in the lectures?? Sometimes I feel like I'm wasting my money and that these channels should profit.
Ikr. I'm currently in high school taking bc calc so I don't have the same struggle as you with feeling like I'm wasting my money, but I can relate to what you're saying though. I never learned this in a classroom because... well school hasn't started quite yet for me. I watched all his vids on derivatives and parametric equations and now I feel like I could teach them to another student because he just does it that well.
i literally pay a thousand dollars to my university and still not understand a thing... but i do get it in a 10 minute video?? you sir are the best, you helped me a lot!
You make everything so easy to understand. I am going into my first calculus course and thought I would be struggling a ton, but if I just make sure to come and learn stuff from your vids while studying everything seems like it will be a breeze.
WOW!!!!! THANK YOU, VERY MUCH, FOR MAKING THIS SO COMPREHENSIBLE!!!! I'M EAGER TO KEEP GOING!!!! THIS IS GREAT!!!! NOW, ...WHERE CAN WE "APPLY" THIS KIND OF MATH???!!!
So clearly explained. No fuss, just , this is how you do it. Thank you for saying it clearly. I have no doubt that in the future we shall come to understand what finding a derivative means in the real world and why it is used but I shall be prepared for that when the time comes. Again a big thank you.
Thanks to your videos I could finally understand 18 hours worth of lectures. My only problem is that I get flustered when I see square roots and fractions. Help me overcome this hahahaha thank you again!
I kept waiting for him to take care of that radical in the denominator -- 1/(2*sqrt(x)) multiply by sqrt(x)/sqrt(x) yielding sqrt(x)/2x -- plug that in much more easily than that first form ;-)
Wow I wish I would have found your channel before my first calculus exam (which the entire class bombed the professor curved it by 50 points). Thank you so much for making these videos you are saving me rn!!
True, the order does not matter in this case, but since the quotient rule (where the order does matter) has f'g as the first term in the numerator, it is best to always start with f'g. Then you'll never run into doubt.
You can often "think backwards" to get these. For example the derivative of what would give us sin ? That would have to be - cos. So we can say that the anti-derivative of sine is negative cosine. :^D
Jhelo Mae Betarmos haha thanks, I haven’t seen this video in ages! I’m doing graduate school mathematics now, hard to believe I still didn’t know the product rule four years ago!!
I've just spent a year doing a maths module with the open University and come out unable to remember any of it. I watch this and it's sticking (and has taken an hour).
Greetings and thank you for all of the great help. Just wondering (around 5:20) why it is d/dx(x) to the exponent of 1/2? I know that the x is originally under a square root, but I'm not sure why that leads to the power of 1/2. Thanks in advance!
The way it works the square root splits everything in half (except for prime numbers). Figuratively speaking, 1 can be cut in half, but in higher math we stick to reality. Although mathematics doesn’t stop. Meaning we imagine the possibilities, not just reality. That is the power of assumption. For that assumption capacity is the source of our mathematics. As you go beyond this point; mathematics becomes a theory. Although this is my theory.
I don't even see how you get that, and I think you want the terms to be distributed. 2e^x( (1/√x) + 2√x ) will work, but that isn't much simpler. To put irrationals on the top of fractions, we can now factor out √x and get 2e^x √x( 1/x + 2).
Why not just expand the brackets by the factor of the function then get the derivative of those terms, instead of finding two different functions of the rule separately then combining them with the primes and normal functions at the end which seems like extra complexity to remember.
You mean to recombine the factors into 3x⁴ + x² and then differentiate it to 12x³ + 2x? Yes, but I suppose we ignore that here in order to teach the rule. The only difference is choosing an expression you couldn't recombine to make it simpler That would require a more complicated outer function that might complicate the teaching process?
As he said, distribute the square root x So, it would be (3x^2 sqr x + 4x sqr x) And then, you'd place the x^3 +2x^2 on top of the fraction that is beside it, so it would be X^3 + 2x^2 ---------------- 2 sqr x You can, then, factor out a x^2 X^2 (x + 2) ---------------- 2 sqr x
can you compute the deriative of G(x)=-3x^5 X (-9x^3). I know the answer is 216x^7 but im stumped when it comes to using the product rule formula. I have half of the equation solved. g'(x)= (-3x^5) X (-27x^2) +(_______) X (______) =216x^7. PLEASEEEE HELPPPP THANK YOU IN ADVANCEEEEEEE
