Question

if f and g are differentiable at x then so is their product. (f.g)'(x)=f(x)g'(x)+g(x)f'(x) lgbasallote plz the answer

Answer 1

this is a fact, not a question

Answer 2

maybe the question is to prove?

Answer 3

depends on what you know how easy this is. it is pretty straightforward to prove from the definition that
\[(f^2)'=2ff'\] then you can write the product as squares and do some minor algebra to get it

Answer 4

yes the prove igbasallote

Answer 5

what ths prove

Answer 6

i need prove lgbasallote

Answer 7

you can start by knowing that
\[(f^2)'=2ff'\] which you can prove by the definition easily.
then note that
\[(f+g)^2=f^2+2fg+g^2\] so
\[fg=\frac{1}{2}((f+g)^2-f^2-g^2)\]

Answer 8

take the derivative of the right hand side using the rule for squares to find the derivative of the left hand side

Answer 9

lgbasallote where r u ?

Answer 10

the proof of this thing is very confusing...but that is as simple as it can be ^

Answer 11

ok

Answer 12

expanding what satellite73 said...

fg = 1/2(2(f+g)(df + dg) - 2fdf -2gdg)

you can cancel the 1/2 there...

so...
fg = (f+g)(df+dg) - (df + dg)
fg =fdf +gdf +fdg +gdg - df - dg

fdf,gdg,df and dg are all equal to 1 if i'm not mistaken...so...

fg = 1 + gdf + fdg + 1 -1 -1

cancel...
fg = gdf + fdg

thus...
the product rule :D

Answer 13

these things are on wikipedia
http://en.wikipedia.org/wiki/Product_rule#Proof_of_the_product_rule