Question

Use the definition of the derivative to prove the formula
(f/g)' = (f'g-g'f)/g^2

Answer 1

its the u/v form

Answer 2

yes

Answer 3

n if u want to check its right or wrong use the f.p.m

Answer 4

So by definition of derivative,
\[ (f/g)'(x) = \lim_{x \rightarrow a} \frac{ (f/g)(x) - (f/g)(a)}{x-a} \]
Are you using this definition, or the one where the limit of delta x or h --> 0?

Answer 5

no not abinatios method ur write james

Answer 6

Lymeng_Chhorn, talk to me. Is what I've written down the definition of derivative you are using or not? Once we have that, I can show you the next steps. But be quick in responding as there are other people's questions out there as well.

Answer 7

the limit of delta --> 0

Answer 8

Ok, so consider the difference quotient
\[ \frac{(f/g)(x + \Delta x) - (f/g)(x)}{\Delta x} \]
This is equal to
\[ \frac{f(x + \Delta x)/g(x+\Delta x)- f(x)/g(x)}{\Delta x} \] by definition of the quotient of functions.

Now we know what we want this to be equal to. So you want to manipulate this expression somehow so you have a limit such as....

Answer 9

\[ \frac{g(x) ( f(x+\Delta x) - f(x) )}{g(x)^2\Delta x} - f(x)\frac{g(x+\Delta x) - g(x)}{g(x)^2 \Delta x}\]

Answer 10

Hence manipulate the difference quotient until you get it in a form like this and then take the limit.

Answer 11

ok i got it :)
Thank @JamesJ