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December 29, 2014

December 29, 2014

Posted by **Karoline** on Saturday, November 21, 2009 at 9:58pm.

i don't get the step where they find the common factor could you help me by doing it step by step please ?

thank you!

- Maths -
**drwls**, Saturday, November 21, 2009 at 10:36pmThere is no such step that i have ever heard of.

Let

u (x) = (5x-2)^3 and

v (x) = (3-x)^6

You want the derivative of u*v with respect to x.

dy/dx = d(uv)/dx = u dv/dx + v du/dx

That is the product rule.

Now all you have to calculate is du/dx and dv/dx, multiply the v and u terms and add the result.

- Maths -
**Karoline**, Sunday, November 22, 2009 at 12:15amthanks!

there is an example here and where i get lost is a step where it goes from

dy/dx = d(uv)/dx = u dv/dx + v du/dx

= (5x-2)^3[-6(3-x)^5]+(3-x)^6[15(5x-2)^2]

= (5x-2)^2(3-x)^5(-30x+12+45-15x)

is there a step between there that you could show me how to do so i can understand it ?

thank you!

- Maths -
**Reiny**, Sunday, November 22, 2009 at 12:29amYou must be talking about the simplication after you found the derivative.

look at the line ....

(5x-2)^3[-6(3-x)^5]+(3-x)^6[15(5x-2)^2]

= -6(5x-2)^3(3-x)^5 + 15(3-x)^6(5x-2)^2

now I see the following common factors

-3 for the constants

(5x-2)^2 and

(3-x)^5

so

= -3(5x-2)^2(3-x)^5[2(5x-2) - 5(3-x)]

= -3(5x-2)^2(3-x)^5[10x - 4 - 15 + 5x]

= -3(5x-2)^2(3-x)^5(15x - 19)

Your last line is not fully factored.

- Maths -
**Karoline**, Sunday, November 22, 2009 at 12:56amah OK, now i get it.

Thank you so much!!

- Maths -
**Karoline**, Sunday, November 22, 2009 at 1:38amhey sorry, one more thing.

How do you find -3

as in

now I see the following common factors

-3

how ?

- Maths -
**MathMate**, Sunday, November 22, 2009 at 10:59am(-30x+12+45-15x)

=-3(15x-19)

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