Question

Find the derivative:

sin(xcosx)

so the answer is after being simplified (cosx-xsinx)cos(xcosx)

I have no idea. how.

Answer 1

this is chain rule

Answer 2

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

Answer 3

ok so:

sin(xcosx)(-sinx)(1)+cos(xcosx)

Answer 4

why is that not the answer I am getting :(

Answer 5

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)

Answer 6

so you end up with

cos(xcos(x)(-xsin(x)+cos(x))

Answer 7

which is the same as

(cos(x)-xsin(x))cos(xcos(x))

Answer 8

does this make sense?

Answer 9

hi