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OpenStudy (anonymous):

Chain rule with e^(x^2)

OpenStudy (anonymous):

e(x^2)*2x

OpenStudy (anonymous):

how did you do it :) ?

OpenStudy (anonymous):

derivative of outside times derivative of inside

OpenStudy (anonymous):

f'(g(x)*g'(x)

OpenStudy (anonymous):

i just dont get it ... what is the outside and what the inside part here ... and how do i figure that out?

OpenStudy (anonymous):

e^x's derivative is just e^x so e^(x^2)'s derivative is just e^(x^2) then you multiply that by the derivative of x^2 which is 2x

OpenStudy (anonymous):

usually just consider the numbers in the parenthesis as the inside when looking at this function (x+2)^3's derivative would be 3(x+2)^2*1

OpenStudy (anonymous):

so it would actually be written "(e^x)^2" ? ... which would make "e^x" the inner function?

OpenStudy (anonymous):

the (x^2) would be the innermost function.

OpenStudy (anonymous):

You have to deal with both e^x^2 function and the x^2 function.

OpenStudy (anonymous):

And according to the chain rule, you multiply both derivatives together. Is this clear?

OpenStudy (anonymous):

x^2 is the inner function but it doesnt matter

OpenStudy (anonymous):

now it is ... thank you soooo much. i really really do appreciate it. i have been pondering over this for over an hour by now.

OpenStudy (anonymous):

you can do derivative of inside times outside or vice versa. Its more on how you define f and g

OpenStudy (anonymous):

try to see if you can distinguish in many problems the outside and inside then you are all set if you can do that because derivatives are easy from there if you know the rules and how to apply them

OpenStudy (anonymous):

Haha, Saif, without a graphing calc, your problem is difficult. Off topic, I've narrowed f'(θ) to -sin(θ) + sin(2θ) = 0 if you can take it from there.

OpenStudy (anonymous):

yea i got it, we arent aloud to use graphing calculators

OpenStudy (anonymous):

its 0 and 2pi

OpenStudy (anonymous):

oopps, I meant -2sin(θ) + sin(2θ) = 0

OpenStudy (anonymous):

Oh nice. Good.

OpenStudy (anonymous):

thanks i gotta get ready, please fan me if i helped

OpenStudy (anonymous):

sorry ... gotta ask once again. how do you know "x^2" is the inner function here?

OpenStudy (anonymous):

its e raised to a parenthesis e^x is the first function and x=x^2

OpenStudy (anonymous):

you have to learn to distinguish between your inside and outside functions

OpenStudy (anonymous):

It is the innermost function because it is the one closest to x. You cannot not differentiate further.

OpenStudy (anonymous):

i really gotta get ready for class so see if IPARKE can help you

OpenStudy (anonymous):

thank you thank you thank you ... big help :)

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