Differentiate using product Rule

**Maths_noob** · June 5th 2009, 08:29 AM

Is This correct?

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

$u=(x+20)^4$

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

$v=(x-30)^3$

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

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

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

**josipive** · June 5th 2009, 08:37 AM

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

**e^(i*pi)** · June 5th 2009, 08:37 AM

Originally Posted by Maths_noob

Is This correct?

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

$u=(x+20)^4$

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

$v=(x-30)^3$

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

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

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

Looks right to me

**Maths_noob** · June 5th 2009, 08:40 AM

Nectar thats lovely jubbly then

Finaly! I know how to use the product rule

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