What's the derivative of 2xsinx and why?

Question

anonymous · Answer

This problem is a combination of 4 formulas in derivatives of a function:
1. d [k f(x)] = k f'(x) dx
2. d (uv) = u dv + v du
3. d [x^n] = n x^(n-1) dx
4. d [sinx] = cos x dx

anonymous · Answer

we can use formula 1... where k=2 and f(x) = xsinx... 
d (2xsinx) = 2 d(xsinx)     (1)

anonymous · Answer

we can use formula 2 in (1)... where u=x and v=sinx...
2 d(xsinx) = x d(sinx) dx + sinx d(x) dx     (2)

anonymous · Answer

we can use formula 3 in u=x... that is du=(1)x^(1-1)=1 dx...
we can use formula 4 in v=sinx... that is dv=cosx dx

anonymous · Answer

rewriting (1) combined with (2)...
d (2xsinx) = 2 (xcosx + sinx) = 2xcosx + 2sinx... answer :)

anonymous · Answer

Ah! Thank you so much!

anonymous · Answer

you're welcome... :)