Question

Anyone good with the f(g(x))g'(x) pattern ?Please help plsss

Answer 1

f(g(x))g'(x) ?
I'm not sure what that is.

Are you referring to the chain rule?\[\Large\rm \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\]

Answer 2

Hi so um cosx/sin^3x u=g(x) what would u equal ?

Answer 3

its like this but this one is hard to follow http://www.webassign.net/ebooks/larsonet5/shell.html?s=170ad96ecdf447814169a8fbe1ffc969&c=285720&f=1073076&type=youbook&id=137&page=333

Answer 4

I can't view YOUR webassign :P

Answer 5

it was an ebook but aw lol

Answer 6

So you have cos x / sin^3x.
What are you trying to do?
Find a derivative?

Answer 7

basically it is written as g(x) multiplied by the derivative of g(x) and then it would ask to find each part, idk how to do that Lol.

Answer 8

@satellite73 ? ;D

Answer 9

Your question isn't clear :(
Maybe take a picture of it and upload with the "Attach File" button.

Answer 10

can u see it..... ? lol

Answer 11

Im sorry ignore the cosx next to u that was my attempt to the problem i forgot to delete it lol

Answer 12

Ah I see now :d
It's a teaching technique for U-substitution.

Answer 13

Ohh lol

Answer 14

please help me with this ? ;D

Answer 15

See if this makes sense to you a sec.
\[\Large\rm \int\limits \frac{1}{[\sin x]^3} \color{orangered}{(\cos x~dx)}=\int\limits \frac{1}{[g(x)]^3}\color{orangered}{g'(x)}\]

Answer 16

ohh so sinx = g(x) ???

Answer 17

\[\Large\rm u=\sin x\]Taking the derivative gives,\[\Large\rm \color{orangered}{du=\cos x~dx}\]

Answer 18

Yah, that's the u that we want :d

Answer 19

hang on im taking it all in im so slow with derivatives lol.

Answer 20

Ohhhh I get it now thanks!!

Answer 21

btw hiii @aum thanks for helping me out earlier with the physics! ;D