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

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- anonymous

- katieb

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- anonymous

this is chain rule

- anonymous

\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)

- anonymous

ok so:
sin(xcosx)(-sinx)(1)+cos(xcosx)

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## More answers

- anonymous

why is that not the answer I am getting :(

- 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

so you end up with
cos(xcos(x)(-xsin(x)+cos(x))

- anonymous

which is the same as
(cos(x)-xsin(x))cos(xcos(x))

- anonymous

does this make sense?

- anonymous

hi |dw:1340953644460:dw|

- 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

\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]

- myininaya

Now use product rule for that one part

- anonymous

OHHHHHHHH

- 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

Sorry @Outkast3r09 I didn't understand the first time you explained that to me.

- 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

then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together

- anonymous

since your u=xcos(x) you have to use product rule to find the derivative

- anonymous

I'm going to try to work it out again

- anonymous

y=sin(xcosx)
y'= cos(xcosx)(x)(-sinx)+cosx
That's the same thing right?
I think I got it.

- anonymous

idek how i got that anymore.

- anonymous

hehe alright start by turning your outer function into y=f(u)

- anonymous

by letting your u be the innre function

- anonymous

ok... let's start from the beginning..
\[\huge y=sin(xcosx) \]
\[\huge y'=cos(xcosx)\cdot [xcosx]' \]
by the chain rule...

- anonymous

i'd start by using the actual theorem dpalnc

- 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

now putting these together...

- anonymous

\[\huge y'=cos(xcosx)\cdot (cosx - xsinx) \]

- anonymous

THANK YOU!!!!!

- anonymous

I got it. 100%

- anonymous

yw...:)

- anonymous

just take it one step at a time and you should be good...:)

- anonymous

:) Alright I will.

- anonymous

:)

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