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C3 differentiation using the product rule

Hey guys, im looking at a question in the textbook and was wondering how they reached that answer.

the question is find dy/dx for y=x(x+2)^2

so we know that u = x, and v = (x+2)^2 . u' = 1 and v' = 2(x+2)

so the general formula is dy/dx = u'v + v'u

now the book says

x*2(x+2) + (x+2)^2 * 1

and here's the strange bit, they then simplify that expression to (x+2)(2x+x+2)

i dont understand how they got from an expression with a + sign to get the expression to multiply, any help guys?
Both have the common factor (x+2) they'll then just go from there I'd assume, try taking (x+2) out (as in at the front like the simplified expression) then working from there. Hope that helps
Reply 2
Avatar for k9markiii
k9markiii
18
Original post by James A
Hey guys, im looking at a question in the textbook and was wondering how they reached that answer.

the question is find dy/dx for y=x(x+2)^2

so we know that u = x, and v = (x+2)^2 . u' = 1 and v' = 2(x+2)

so the general formula is dy/dx = u'v + v'u

now the book says

x*2(x+2) + (x+2)^2 * 1

and here's the strange bit, they then simplify that expression to (x+2)(2x+x+2)

i dont understand how they got from an expression with a + sign to get the expression to multiply, any help guys?


(x+2) is in both parts of the equation.
2x(x+2)+(x+2)^2
2x(x+2)+(x+2)(x+2)
(x+2)(2x+x+2)
(x+2)(3x+2)
Reply 3
Avatar for F1's Finest
F1's Finest
OP
21
Original post by geordiecoinman
Both have the common factor (x+2) they'll then just go from there I'd assume, try taking (x+2) out (as in at the front like the simplified expression) then working from there. Hope that helps



Original post by k9markiii
(x+2) is in both parts of the equation.
2x(x+2)+(x+2)^2
2x(x+2)+(x+2)(x+2)
(x+2)(2x+x+2)
(x+2)(3x+2)


cheers guys, i cant believe i forgot about them having a common factor!! :colondollar:

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