SandyHearts
Find the derivative:
sin(xcosx)
so the answer is after being simplified (cosx-xsinx)cos(xcosx)
I have no idea. how.
Delete
Share
This Question is Closed
Outkast3r09
Best Response
You've already chosen the best response.
1
this is chain rule
Outkast3r09
Best Response
You'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)
SandyHearts
Best Response
You've already chosen the best response.
0
ok so:
sin(xcosx)(-sinx)(1)+cos(xcosx)
SandyHearts
Best Response
You've already chosen the best response.
0
why is that not the answer I am getting :(
Outkast3r09
Best Response
You'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)
Outkast3r09
Best Response
You've already chosen the best response.
1
so you end up with
cos(xcos(x)(-xsin(x)+cos(x))
Outkast3r09
Best Response
You've already chosen the best response.
1
which is the same as
(cos(x)-xsin(x))cos(xcos(x))
Outkast3r09
Best Response
You've already chosen the best response.
1
does this make sense?
meera_yadav
Best Response
You've already chosen the best response.
1
hi |dw:1340953644460:dw|
SandyHearts
Best Response
You'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
myininaya
Best Response
You've already chosen the best response.
0
\[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]
myininaya
Best Response
You've already chosen the best response.
0
Now use product rule for that one part
SandyHearts
Best Response
You've already chosen the best response.
0
OHHHHHHHH
SandyHearts
Best Response
You've already chosen the best response.
0
>.<
SandyHearts
Best Response
You'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.
SandyHearts
Best Response
You've already chosen the best response.
0
Sorry @Outkast3r09 I didn't understand the first time you explained that to me.
Outkast3r09
Best Response
You'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
Outkast3r09
Best Response
You'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
Outkast3r09
Best Response
You've already chosen the best response.
1
since your u=xcos(x) you have to use product rule to find the derivative
SandyHearts
Best Response
You've already chosen the best response.
0
I'm going to try to work it out again
SandyHearts
Best Response
You'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.
SandyHearts
Best Response
You've already chosen the best response.
0
idek how i got that anymore.
Outkast3r09
Best Response
You've already chosen the best response.
1
hehe alright start by turning your outer function into y=f(u)
Outkast3r09
Best Response
You've already chosen the best response.
1
by letting your u be the innre function
dpaInc
Best Response
You'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...
Outkast3r09
Best Response
You've already chosen the best response.
1
i'd start by using the actual theorem dpalnc
dpaInc
Best Response
You'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...
dpaInc
Best Response
You've already chosen the best response.
1
now putting these together...
dpaInc
Best Response
You've already chosen the best response.
1
\[\huge y'=cos(xcosx)\cdot (cosx - xsinx) \]
SandyHearts
Best Response
You've already chosen the best response.
0
THANK YOU!!!!!
SandyHearts
Best Response
You've already chosen the best response.
0
I got it. 100%
dpaInc
Best Response
You've already chosen the best response.
1
yw...:)
dpaInc
Best Response
You've already chosen the best response.
1
just take it one step at a time and you should be good...:)
SandyHearts
Best Response
You've already chosen the best response.
0
:) Alright I will.
dpaInc
Best Response
You've already chosen the best response.
1
:)