Question

I am asked to find the differential at dy for the following function.

y=xcos2x

Can someone please get me started on this problem, I do not understand what I am being asked.

thanks...

Answer 1

product rule for this one
\[(fg)'=f'g+g'f\] with
\[f(x)=x,f'(x)=1,g(x)=\cos(2x),g'(x)=-2\sin(2x)\]

Answer 2

I looked in the book and found this...

dy=f'(x)dx

But I do not know how this relates to my problem...

Answer 3

product rule...
d/dx (u.v) = u. d/dx (v) + v. d/dx (u)

Answer 4

u= x and v = cos2x put in above equation..

Answer 5

Do I just do the product rule and then multiply the result of the product rule by dx

Answer 6

you'll get..
x d/dx (cos2x) + cos2x d/dx (x)

Answer 7

can you solve it now..??