anonymous
  • anonymous
Find the derivative: sin(xcosx) so the answer is after being simplified (cosx-xsinx)cos(xcosx) I have no idea. how.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
this is chain rule
anonymous
  • anonymous
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)
anonymous
  • anonymous
ok so: sin(xcosx)(-sinx)(1)+cos(xcosx)

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anonymous
  • anonymous
why is that not the answer I am getting :(
anonymous
  • anonymous
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)
anonymous
  • anonymous
so you end up with cos(xcos(x)(-xsin(x)+cos(x))
anonymous
  • anonymous
which is the same as (cos(x)-xsin(x))cos(xcos(x))
anonymous
  • anonymous
does this make sense?
anonymous
  • anonymous
hi |dw:1340953644460:dw|
anonymous
  • anonymous
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
myininaya
  • myininaya
\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]
myininaya
  • myininaya
Now use product rule for that one part
anonymous
  • anonymous
OHHHHHHHH
anonymous
  • anonymous
>.<
anonymous
  • anonymous
I see @Outkast3r09 tried explaining that to me but I was really not understanding all the notations you used. Thank you all so much.
anonymous
  • anonymous
Sorry @Outkast3r09 I didn't understand the first time you explained that to me.
anonymous
  • anonymous
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
anonymous
  • anonymous
then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together
anonymous
  • anonymous
since your u=xcos(x) you have to use product rule to find the derivative
anonymous
  • anonymous
I'm going to try to work it out again
anonymous
  • anonymous
y=sin(xcosx) y'= cos(xcosx)(x)(-sinx)+cosx That's the same thing right? I think I got it.
anonymous
  • anonymous
idek how i got that anymore.
anonymous
  • anonymous
hehe alright start by turning your outer function into y=f(u)
anonymous
  • anonymous
by letting your u be the innre function
anonymous
  • anonymous
ok... let's start from the beginning.. \[\huge y=sin(xcosx) \] \[\huge y'=cos(xcosx)\cdot [xcosx]' \] by the chain rule...
anonymous
  • anonymous
i'd start by using the actual theorem dpalnc
anonymous
  • anonymous
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...
anonymous
  • anonymous
now putting these together...
anonymous
  • anonymous
\[\huge y'=cos(xcosx)\cdot (cosx - xsinx) \]
anonymous
  • anonymous
THANK YOU!!!!!
anonymous
  • anonymous
I got it. 100%
anonymous
  • anonymous
yw...:)
anonymous
  • anonymous
just take it one step at a time and you should be good...:)
anonymous
  • anonymous
:) Alright I will.
anonymous
  • anonymous
:)

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