More Chain Rule (NancyPi)
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- Опубликовано: 22 окт 2019
- MIT grad shows how to use the chain rule for EXPONENTIAL, LOG, and ROOT forms and how to use the chain rule with the PRODUCT RULE to find the derivative. To skip ahead: 4) For an example with an EXPONENTIAL function that needs the chain rule to take the derivative, skip to time 0:32. 5) For an example with a natural LOG, skip to 2:34. 6) For an example with a SQUARE ROOT, skip to 4:41. 7) For how to use the CHAIN RULE with the PRODUCT RULE, and how to know which to use first, skip to 7:17. For an introduction of HOW and WHEN to use the chain rule, jump to my FIRST chain rule video: • The Chain Rule... How?... . Nancy formerly of MathBFF explains the steps.
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The CHAIN RULE is one of the derivative rules. You need it to take the derivative when you have a function inside a function, or a "composite function". This video picks up at the end of the first chain rule video ("Chain Rule...How? When?" at • The Chain Rule... How?... ) and shows more examples of using the chain rule, as well as how to combine the chain rule with the product rule.
For an intro of how to take derivatives, jump to: • Derivatives... How? (N...
4) EXPONENTIAL: for the example y = e^(3x), first take the derivative of the outside function (the exponential function, e). The derivative of the exponential function is just the exponential function itself. When writing the outside derivative, leave the inside function (3x) the same, so write e^(3x). Then by the chain rule, you multiply by the derivative of just the inner function (3x). Since the inside derivative is 3, the answer for the full derivative dy/dx is 3e^(3x).
5) NATURAL LOG example: To find the derivative of y = ln(5x), use the same steps as above to first take the outside derivative and then multiply by the inside derivative. 5x is inside the natural log. The outer function here is the natural log, ln. To take the derivative of the natural log of something, we write 1 over that something, so 1 over 5x. We leave the inside function alone here when writing the outside derivative, so we write 1/(5x). Remember that we're not done yet. We also need to multiply by the derivative of the inside function. The derivative of 5x is just 5, so our final simplified answer for dy/dx is 1/x.
6) SQUARE ROOT example: For this example, y = sqrt(x^2 + 1), the x^2 + 1 is under a square root. The square root is the outside, outer function. When you have a root of any kind, before you take the derivative, it's easiest to rewrite it in the form of a fractional power, so that you can use the POWER RULE when taking its derivative. The outside function here is then the 1/2 power, and the inside function is x^2 + 1. To take the derivative of the outside one-half power, we use the power rule to bring down the 1/2 power to the front and reduce the old 1/2 power by 1. We then multiply by the inside derivative, and after simplifying get a derivative of y' = x / sqrt(x^2 + 1).
Here I used Lagrange notation (y') for the derivative instead of the Liebniz notation from before (dy/dx). To learn the POWER RULE, jump to my video at: • Power Rule... How? (Na...
7) CHAIN RULE WITH THE PRODUCT RULE: Sometimes you might need to use the chain rule combined with the product rule. How do you know whether to apply the chain rule first, or the product rule first? In this example, y = x^3 (2x - 5)^4, although the second factor here will need the chain rule for its derivative, overall in the biggest view of this function, we have a product of two factors. So we will use the product rule first here on the two factors, and then use the chain rule inside the product rule. However, for an example like ln[x^3 (2x-5)^4)], you would need to apply the chain rule first on the ln form, and then since your inside function is a product, you would need the product rule after that. Basically, the first thing you'll do is whatever you need to use to handle the outermost form of your function.
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Editor: Miriam Nielsen of zentouro
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Kshitish Sorte 3:54
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This and the first chain rule video really helped me understand derivatives and helped with integrals. I've been stuck on this for so long. Thank you.
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1:22 having the video speedup for "e" by itself. lol ! Also welcome back. :)
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I did not comprehend the chain rule at all until watching your videos. 3 weeks into calculus, fresh out of precalc, after watching your videos for precalc... I was dumb to think I could get through calculus without watching your videos lol. Thank you. You're a wizard!
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If someone sees this what happens to the power of 4 in this step: 8x^3(2x-5)^3+3x^2(2x-5)^4, I get how x^2 was a common factor leading to [8x+3(2x-5)] but why Is (2x-5) no longer raised to the power of 4
I'm with you. Didn't quite get that last simplification. Nor did she refer us to another one of her vids. I guess it is just basic algebra and so we are SOL.
Hi Lalo .g17 and fly4doe, just like x^2 being a common factor so is (2x-5). In fact you could factor up to (2x-5)^3 from both terms. It will leave this..
y'= x^2(2x-5)^3
@@eg8charlesthanks alot that was helpful❤
Hello and thank you, Nancy. I'm new to derivative and the chain rule so this helped me a lot!
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Very nice video! Thanks Nancy!
Just a quick note about the result in 3:49
ln(k*x) = ln(k) + ln(x) (property of logs)
d/dx(ln(kx)) = 0 + 1/x = 1/x (k is a constant)
When I went through this result I found it very interesting, so thanks a lot for making me more curious!
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YAY! The internet's best math pedagogue has returned! Welcome back!
Pedagogue; A teacher, especially a strict or pedantic one.
Jeremiah Diaz The word is used as a term of art among teachers to refer to someone who is skilled and experienced in that field. Remember that dictionaries don't always tell you the whole story.
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Here's an exercise you could use to demonstrate both the chain rule and the product rule: finding the derivative of x**x!
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