# Maths

posted by on .

Differentiate y=(5x-2)^3(3-x)^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 - ,

There 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 - ,

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
= (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 - ,

You 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 - ,

ah OK, now i get it.
Thank you so much!!

• Maths - ,

hey sorry, one more thing.
How do you find -3

as in
now I see the following common factors
-3
how ?

• Maths - ,

(-30x+12+45-15x)
=-3(15x-19)