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SandyHearts
Group Title
Find the derivative:
sin(xcosx)
so the answer is after being simplified (cosxxsinx)cos(xcosx)
I have no idea. how.
 2 years ago
 2 years ago
SandyHearts Group Title
Find the derivative: sin(xcosx) so the answer is after being simplified (cosxxsinx)cos(xcosx) I have no idea. how.
 2 years ago
 2 years ago

This Question is Closed

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
this is chain rule
 2 years ago

Outkast3r09 Group TitleBest 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)
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
ok so: sin(xcosx)(sinx)(1)+cos(xcosx)
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
why is that not the answer I am getting :(
 2 years ago

Outkast3r09 Group TitleBest 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)
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
so you end up with cos(xcos(x)(xsin(x)+cos(x))
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
which is the same as (cos(x)xsin(x))cos(xcos(x))
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
does this make sense?
 2 years ago

meera_yadav Group TitleBest ResponseYou've already chosen the best response.1
hi dw:1340953644460:dw
 2 years ago

SandyHearts Group TitleBest 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
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
Now use product rule for that one part
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
OHHHHHHHH
 2 years ago

SandyHearts Group TitleBest 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.
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
Sorry @Outkast3r09 I didn't understand the first time you explained that to me.
 2 years ago

Outkast3r09 Group TitleBest 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
 2 years ago

Outkast3r09 Group TitleBest 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
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
since your u=xcos(x) you have to use product rule to find the derivative
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
I'm going to try to work it out again
 2 years ago

SandyHearts Group TitleBest 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.
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
idek how i got that anymore.
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
hehe alright start by turning your outer function into y=f(u)
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
by letting your u be the innre function
 2 years ago

dpaInc Group TitleBest 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...
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
i'd start by using the actual theorem dpalnc
 2 years ago

dpaInc Group TitleBest 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...
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.1
now putting these together...
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.1
\[\huge y'=cos(xcosx)\cdot (cosx  xsinx) \]
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
THANK YOU!!!!!
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
I got it. 100%
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.1
just take it one step at a time and you should be good...:)
 2 years ago

SandyHearts Group TitleBest ResponseYou've already chosen the best response.0
:) Alright I will.
 2 years ago
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