Product rule..(x^3-1) (x+2) for calculus...i have been getting X^4 + 3X^2 + 2X...?

Question

but the answer in the book is 4x^3+ 3x^2-1

I have no clue how to get to the right anwer..please help

Answer 1

(x³ - 1)(x + 1) = u*v
d(u*v) = u'v + v'u
=(3x²)(x+1) + (1)(x² - 1)
= 3x³ + 3x² - 1

Answer 2

I think what the question wants you to do is to differentiate the function (x^3-1) (x+2) using the product rule.

I don't know how you got that expression, but if you were to expand the factors you shold get something like below:

(x^3-1) (x+2) = x^4 + 2x^3 - x - 2

However, it wants you to use the product rule, so you shouldn't expand the factors but differentiate it straightaway.

The formula is
f(x) = uv, f '(x) = u*dv + du*v

so, i'll call f(x) = (x^3-1) (x+2)

hence,

f '(x) = (x^3 - 1)(1) + 3x^2 * (x+2)
= x^3 - 1 + 3x^3 + 6x^2
= 4x^3 + 6x^2 - 1

I think the answer in you book has a typo where the term 3x^2 should actually be 6x^2.

Hope this helps! =)

Answer 3

(x^3-1)(1) +(x+2)(3x^2)
= x^3-1 +3x^3 +6x^2
= 4x^3 +6x^2-1
I believe the book to be wrong.

Answer 4

u = x^3 - 1

du = 3x^2

v = x + 2

dv = 1

d(uv) = udv + vdu

=>(x^3-1)(1) + (x+2)(3x^2)

=>x^3 - 1 + 3x^3 + 6x^2

=>4x^3 + 6x^2 - 1

Answer 5

It's first times derivative second times second times derivative first,

so

3x^2(x+2) + x^3 -1
3x^3+6x^2+x^3-1
4x^3+6x^2-1

So I suppose your book is wrong too.