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Find the derivative:
sin(xcosx)
so the answer is after being simplified (cosxxsinx)cos(xcosx)
I have no idea. how.
 one year ago
 one year ago
Find the derivative: sin(xcosx) so the answer is after being simplified (cosxxsinx)cos(xcosx) I have no idea. how.
 one year ago
 one year ago

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Outkast3r09Best ResponseYou've already chosen the best response.1
this is chain rule
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
ok so: sin(xcosx)(sinx)(1)+cos(xcosx)
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
why is that not the answer I am getting :(
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
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)
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so you end up with cos(xcos(x)(xsin(x)+cos(x))
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
which is the same as (cos(x)xsin(x))cos(xcos(x))
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
does this make sense?
 one year ago

meera_yadavBest ResponseYou've already chosen the best response.1
hi dw:1340953644460:dw
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
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
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
Now use product rule for that one part
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
I see @Outkast3r09 tried explaining that to me but I was really not understanding all the notations you used. Thank you all so much.
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
Sorry @Outkast3r09 I didn't understand the first time you explained that to me.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
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
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
since your u=xcos(x) you have to use product rule to find the derivative
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
I'm going to try to work it out again
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
y=sin(xcosx) y'= cos(xcosx)(x)(sinx)+cosx That's the same thing right? I think I got it.
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
idek how i got that anymore.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
hehe alright start by turning your outer function into y=f(u)
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
by letting your u be the innre function
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
ok... let's start from the beginning.. \[\huge y=sin(xcosx) \] \[\huge y'=cos(xcosx)\cdot [xcosx]' \] by the chain rule...
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
i'd start by using the actual theorem dpalnc
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
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...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
now putting these together...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
\[\huge y'=cos(xcosx)\cdot (cosx  xsinx) \]
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
just take it one step at a time and you should be good...:)
 one year ago

SandyHeartsBest ResponseYou've already chosen the best response.0
:) Alright I will.
 one year ago
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