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this is chain rule

\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)

ok so:
sin(xcosx)(-sinx)(1)+cos(xcosx)

why is that not the answer I am getting :(

so you end up with
cos(xcos(x)(-xsin(x)+cos(x))

which is the same as
(cos(x)-xsin(x))cos(xcos(x))

does this make sense?

hi |dw:1340953644460:dw|

\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]

Now use product rule for that one part

OHHHHHHHH

>.<

Sorry @Outkast3r09 I didn't understand the first time you explained that to me.

since your u=xcos(x) you have to use product rule to find the derivative

I'm going to try to work it out again

y=sin(xcosx)
y'= cos(xcosx)(x)(-sinx)+cosx
That's the same thing right?
I think I got it.

idek how i got that anymore.

hehe alright start by turning your outer function into y=f(u)

by letting your u be the innre function

i'd start by using the actual theorem dpalnc

now putting these together...

\[\huge y'=cos(xcosx)\cdot (cosx - xsinx) \]

THANK YOU!!!!!

I got it. 100%

yw...:)

just take it one step at a time and you should be good...:)

:) Alright I will.

:)