SOLUTION: Evening Everyone. Been a while since i attempeted anything like this and trying to brush up a little, i have attempted to anser the first two parts if somebody would be so kin

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Question 374974: Evening Everyone.
Been a while since i attempeted anything like this and trying to brush up a little, i have attempted to anser the first two parts if somebody would be so kind as to check them and assist on the third part

Use the chain rule to find the derivatives of the following functions:
(a) y = e^-x2
I Got = -2e^(-x^2)
(b) y = cos(3x^2 + 2x)
I Got = (-Sin).(3x^2+2x).(6x+2)
(c) y = sin^2(3x^2 - 1)
Thank You in antiscipation
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a) No,
Let u=x%5E2, du%2Fdx=2x, dy%2Fdu=e%5Eu
dy%2Fdx=%28dy%2Fdu%29%28du%2Fdx%29
dy%2Fdx=e%5Eu%2A2x
dy%2Fdx=2xe%5E%28x%5E2%29
.
.
.
b) Yes, dy%2Fdx=-2%283x%2B1%29sin%283x%5E2%2B2x%29
.
.
.
c) You didn't provide an answer.
Let u=sin%283x%5E2-1%29, y=u%5E2, dy%2Fdu=2u
Let v=3x%5E2-1.
u=sin%28v%29
du%2Fdv=cos%28v%29
dv%2Fdx=6x
.
.
.
dy%2Fdx=%28dy%2Fdu%29%2A%28du%2Fdv%29%2A%28dv%2Fdx%29
dy%2Fdx=2u%2Acos%28v%29%2A6x
highlight%28dy%2Fdx=12x%2Asin%283x%5E2-1%29%2Acos%283x%5E2-1%29%29

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


a)

Let , then





Hence:



Part b you got spot on.

Part c just requires you to extend the chain rule, which can be done indefinately. So, instead of



you need to use:



For:



Let , hence

Let , hence

And , hence

Putting it all together:



By the way, the rule is one question per post. I let you slide because you had attempted two of your three problems.

John

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