If anyone knows how to do this that would be helpful. Thank you.
2007-05-20 05:09:46 · 8 answers · asked by Anna D 1 in Science & Mathematics ➔ Mathematics
Follow 2 rules
1. The derivative of any polynomial y=ax^n is y'=n*ax^(n-1)
2. The derivative of the sum of two terms f(x)=g(x)+h(x) is the sum of the derivative of both terms f'(x)=g'(x)+h'(x)
so...
f(x) = 4x^3 - 4x
f'(x) = 3*4x^(3-1) - 1*4x^(1-1)
f'(x) = 12x^2-4
Assuming you want to find the derivative at x=-3
f'(3) = 12*(-3^2) - 4
f'(3) = 12*9 - 4
f'(3) = 108 - 4
f'(3) = 104
2007-05-20 05:16:14 · answer #1 · answered by Astral Walker 7 · 0⤊ 0⤋
Bring the exponent down from the variable you are taking a derivative from. If a number doesnt have the derivative then it will become 0. When you bring that exponent down multiply it by the number in front of the variable. Minus 1 fromthe number in the exponent and it becomes the new exponent.
4x^3-4x=
(4*3)x^(3-1)=12x^2
(4*1)x^(1-1)=4x^0=4
12x^2-4 is the derivative of your equation. If x=-3 then plug it in and solve.
(12)(9)-4=104
2007-05-20 05:16:35 · answer #2 · answered by Cool Nerd At Your Service 4 · 0⤊ 0⤋
To take the derivative of a term, subtract the power by 1 and multiply the coefficient in front of that term by the initial power.
So: f'(x) = 12x^2 - 4
Now plug in -3 for x
f'(-3) = 12 ( -3)^2 -4 = 12*9 - 4 = 104
2007-05-20 05:13:01 · answer #3 · answered by Lilovacookedrice 3 · 0⤊ 0⤋
The derivative is
f'(x) = 12x^2 - 4
using the general power rule.
If you want the tangent line at x=-3, plug in -3 for f(x) to get the point P(-3, -96).
Plug in -3 into f'(x) to get the slope of 104. You have everything you need now, so write the equation of the tangent line in point-slope form.
y+96=104(x+3)
2007-05-20 05:17:19 · answer #4 · answered by jsoos 3 · 0⤊ 0⤋
To find the derative in this example, you need to know the power reduction formula, which is:
the derivative of x^n is n*x^(n-1).
Using the power reduction formula, you get:
f`(x) = 12x^2 - 4
f`(-3) = 12(-3)^2 - 4
f`(-3) = 108 - 4
f1(-3) = 104
hope this helps
2007-05-20 05:16:24 · answer #5 · answered by bballl 2 · 0⤊ 0⤋
f(x) = 4x^3 -4x @ x= -3
f'(x) = (4)(3)x^(3-1)-4(1)x^(1-1) (using the power rule)
f'(x) = 12x^2 -4
f'(-3) = 12(-3)^2 - 4 = 104
2007-05-20 05:15:25 · answer #6 · answered by hawkeye3772 4 · 0⤊ 0⤋
when f(x) = 4x^3 - 4x
f'(x) = 12x^2 - 4
f'(-3) = 12(-3)^2 - 4 = 104
2007-05-20 05:23:03 · answer #7 · answered by d_latha 1 · 0⤊ 0⤋
use result:-
f (x) = a.x^n
f `(x) = n.a.x^(n - 1)
f `(x) = 12.x² - 4
f `(- 3) = 108 - 4
f `(- 3) = 104
2007-05-20 08:25:40 · answer #8 · answered by Como 7 · 0⤊ 0⤋