Posted by Karoline on Saturday, November 21, 2009 at 9:58pm.
Differentiate y=(5x2)^3(3x)^6 using the product rule.
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 10:36pm
There is no such step that i have ever heard of.
Let
u (x) = (5x2)^3 and
v (x) = (3x)^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 12:15am
thanks!
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
= (5x2)^3[6(3x)^5]+(3x)^6[15(5x2)^2]
= (5x2)^2(3x)^5(30x+12+4515x)
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 12:29am
You must be talking about the simplication after you found the derivative.
look at the line ....
(5x2)^3[6(3x)^5]+(3x)^6[15(5x2)^2]
= 6(5x2)^3(3x)^5 + 15(3x)^6(5x2)^2
now I see the following common factors
3 for the constants
(5x2)^2 and
(3x)^5
so
= 3(5x2)^2(3x)^5[2(5x2)  5(3x)]
= 3(5x2)^2(3x)^5[10x  4  15 + 5x]
= 3(5x2)^2(3x)^5(15x  19)
Your last line is not fully factored.

Maths  Karoline, Sunday, November 22, 2009 at 12:56am
ah OK, now i get it.
Thank you so much!!

Maths  Karoline, Sunday, November 22, 2009 at 1:38am
hey 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+4515x)
=3(15x19)
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