g(x)=(x^3-7)(2x^2+3) then g'(-1)=?

Answer 1

The easiest thing to to do here is to multiply out, and then take the derivative. So,

g(x) = 2x^5+3x^3-14x^2-21
So, g'(x) = 10x^4+9x^2-28x
So g'(-1) = 10(-1)^4 + 9(-1)^2 - (28)(-1) = 10+9+28 = 47

Answer 2

First, find g(x):

g(x)=2x^5+3x^3-14x^2-21.

g'(x) is the FIRST DERIVATIVE of g(x), thus:

g'(x)=2*5 x^4 + 3*3 x^2 -14*2 x +0
=10x^4 + 9x^2 -28x

(derivative of ax^n = a*n x^(n-1), derivative of constant=0).

plug in -1 to g'(x):

g'(-1) = 10(-1)^4 + 9(-1)^2 -28(-1)
= 10+9+28
= 47.

Answer 3

g(x)=(x^3-7)(2x^2+3) = 2x^5 + 3x^3 - 14x^2 - 21

g'(x) = 10x^4 + 9x^2 - 28x

g'(-1) = 10 + 9 + 28 = 47

Answer 4

g `(x) = (3x²).(2x² + 3) + (4x).(x³ - 7)
g `(x) = 6 x^4 + 9 x² + 4 x^4 - 28 x
g `(x) = 10 x^4 + 9 x² - 28 x
g `(-1) = 10 + 9 + 28
g `(- 1) = 47

Answer 5

I am lazy, so i will use the product rule:
(ef)'=e'f+ef'
g(x)=(x^3-7)(2x^2+3)
e=(x^3-7)
f=(2x^2+3)
g'=(3x^2)(2x^2+3)+(x^3-7)(4x)
g'=(3)(5)+(-8)(-4)
g'=15+32=47

Answer 6

this question was just asked by someone else - you guys should get together and have a study group:)

see a couple of questions ago