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

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so: sin(xcosx)(sinx)(1)+cos(xcosx)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is that not the answer I am getting :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright 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
 3 years ago
Best ResponseYou've already chosen the best response.0so you end up with cos(xcos(x)(xsin(x)+cos(x))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which is the same as (cos(x)xsin(x))cos(xcos(x))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does this make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi dw:1340953644460:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry @Outkast3r09 I didn't understand the first time you explained that to me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it 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
 3 years ago
Best ResponseYou've already chosen the best response.0then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since your u=xcos(x) you have to use product rule to find the derivative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm going to try to work it out again

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0idek how i got that anymore.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hehe alright start by turning your outer function into y=f(u)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by letting your u be the innre function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok... let's start from the beginning.. \[\huge y=sin(xcosx) \] \[\huge y'=cos(xcosx)\cdot [xcosx]' \] by the chain rule...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'd start by using the actual theorem dpalnc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now... \[\huge [xcosx]'=x'\cdot cosx + x \cdot [cosx]' \] \[\huge [xcosx]'=1\cdot cosx + x \cdot (sinx) \] \[\huge [xcosx]'= cosx  x sinx \] by the product rul...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now putting these together...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge y'=cos(xcosx)\cdot (cosx  xsinx) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just take it one step at a time and you should be good...:)
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