Find the derivative of y= xsinx+ cos x

Question

anonymous · Answer

Please don't just give me the answer and actually explain it. thank you

raden · Answer

for xsinx, use this rule :
(uv)' = u'v + uv'

raden · Answer

so, what is derivative of xsinx ?
let u = x and v = sinx
so,
(xsinx)' = 1 * sinx + xcosx = sinx + xcosx

anonymous · Answer

wait how did you get that? I got 1 * cos x + sinx* x

raden · Answer

u = x ----> u' = 1
v = sinx ----> v' = cosx

(uv)' = u'v + uv'

 (xsinx)' = 1 * sinx + xcosx = sinx + xcosx

anonymous · Answer

ohh okay. So then what?

raden · Answer

btw, the question is for y = xsinx+ cos x
derivative of xsinx already we get it, the rest is derivative of cos x = -sinx

therefore,
if y=xsinx+ cos x then
y' = sinx + xcosx - sinx = xcosx

anonymous · Answer

So then wait, y'= xcosx because the sin's cancel out?

raden · Answer

yes, sinx-sinx=0

anonymous · Answer

oh okay! thank you so much!

raden · Answer

you're welcome