Prove please?

Question

Prove that
when u = f(x) and v = g(x)
d/dx (uv) = v du/dx + u dv/dx

^_^

Answer 1

All you needed to do was google it...

http://tutorial.math.lamar.edu/AllBrowsers/2413/ProductRuleProof.asp

Answer 2

its a product rule

see
if u can remember d/dx(u(x)v(x))=lim h->0 {u(x+h)v(x+h)-u(x)v(x)}/h

then if we add 0 in the form of u(x+h)v(x) -u(x+h)v(x) to the numerator, and after performing some minor algebra

d/dx(u(x)v(x))=lim h->0 {u(x+h)v(x+h)+u(x+h)v(x)-u(x+h)v(x)-u(x)v(x)}/h

=lim h->0 {u(x+h){[v(x+h)-v(x)]+v(x)[u(x+h)-u(x)]}}/h

we have du(x)/dx=lim h->0 {u(x+h)-u(x)}/h

and dv(x)/dx=lim h->0 {v(x+h)-v(x)}/h


also lim h->0 u(x+h)=u(x)

lim h->0 v(x+h)=v(x)


it proves
d/dx(u(x)v(x))=u(x)dv/dx+v(x)du/dx

so its proved

Answer 3

Use the definition for the derivative:
d/dx f(x)= lim(h->0) (f(x+h)-f(x))/h.

In this case d/dx (fg) = lim (f(x+h) g(x+h) - f(x) g(x) )/h

Replace f(x+h) by f(x)+h f´(x) + O(h^2), same for g(x).

Simplify.