# Thread: Differentiate using product Rule

1. ## Differentiate using product Rule

Is This correct?

$\displaystyle (x+20)^4 (x-30)^3$

$\displaystyle u=(x+20)^4$

$\displaystyle du/dx = 4(x+20)^3$

$\displaystyle v=(x-30)^3$

$\displaystyle dv/dx=3(x-30)^2$

$\displaystyle dy/dx=v.du/dx + u . dv/dx$

$\displaystyle dy/dx = (x-30)^3 . 4 (x+20)^3 + (x+20)^4 . 3(x-30)^2$

2. yes

( f( x )*g( x ) ) ' = f ' ( x )*g( x ) + g ' ( x )*f( x )

you can put:

if f( x ) = ( x + 20 )^4 => f ' ( x ) = 4*( x + 20 )^3

g( x ) = ( x - 30 )^3 => g ' ( x ) = 3*( x - 30 )^2

and if I use that in the formula above i will get the same thing as you

3. Originally Posted by Maths_noob
Is This correct?

$\displaystyle (x+20)^4 (x-30)^3$

$\displaystyle u=(x+20)^4$

$\displaystyle du/dx = 4(x+20)^3$

$\displaystyle v=(x-30)^3$

$\displaystyle dv/dx=3(x-30)^2$

$\displaystyle dy/dx=v.du/dx + u . dv/dx$

$\displaystyle dy/dx = (x-30)^3 . 4 (x+20)^3 + (x+20)^4 . 3(x-30)^2$
Looks right to me

4. Nectar thats lovely jubbly then
Finaly! I know how to use the product rule