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Find the derivative: sin(xcosx) so the answer is after being simplified (cosx-xsinx)cos(xcosx) I have no idea. how.

Mathematics
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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)

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Other answers:

why is that not the answer I am getting :(
alright so derivative of sin(u) = cos(u) so \[\frac{dy}{du}=cos(u)\] next do the product rule with your u=xcos(x) u'=vw'+wv'=-xsin(x)+cos(x)
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|
from this cos(xcos(x)(-xsin(x)+cos(x)) all i can get is cos(xcos(x)(-xsin(x)) where does the +cos(x) come from
\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]
Now use product rule for that one part
OHHHHHHHH
>.<
I see @Outkast3r09 tried explaining that to me but I was really not understanding all the notations you used. Thank you all so much.
Sorry @Outkast3r09 I didn't understand the first time you explained that to me.
it comes from the chain rule do you understand this \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] if you name your inner function u and your outer function y, your outer function will become y=f(u).. so now you find the derivative of your y in respects to u
then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together
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
ok... let's start from the beginning.. \[\huge y=sin(xcosx) \] \[\huge y'=cos(xcosx)\cdot [xcosx]' \] by the chain rule...
i'd start by using the actual theorem dpalnc
now... \[\huge [xcosx]'=x'\cdot cosx + x \cdot [cosx]' \] \[\huge [xcosx]'=1\cdot cosx + x \cdot (-sinx) \] \[\huge [xcosx]'= cosx - x sinx \] by the product rul...
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.
:)

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