# Differentiate using product Rule

• Jun 5th 2009, 08:29 AM
Maths_noob
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\$
• Jun 5th 2009, 08:37 AM
josipive
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
• Jun 5th 2009, 08:37 AM
e^(i*pi)
Quote:

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 (Smile)
• Jun 5th 2009, 08:40 AM
Maths_noob
Nectar thats lovely jubbly then :)
Finaly! I know how to use the product rule