Calculus, finding g'''(3)!?

Question

Hey,

Ok so the question goes like this:
if g(x) = 1 / 2x - 4, find g'''(3)?

How would I find the derivative of the derivative! When I tried both power rule and the quotient rule I do not get the right answer! Which by the way is supposed to be -3! Can someone help out please!

Answer 1

you just keep on deriving.

I find it easier to split things into exponents.

I am assuming your question goes

(1/(2x)) - 4 = g(x)
or

(1/2)*x^-1 - 4 = g(x)

differentiate once

g'(x) = -(1/2) * x^-2
g''(x) = -2 * (-1/2) * x^-3 = x^-3
g'''(x) = -3 * x^-4 = -3/(x^-4) **ans**

Plug 3 into this... bah

For reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).


===!using the correct g(x) which should be g(x) =1 / (2x -4)===

g(x) = 1 / (2x -4) = (2x - 4)^-1 don't forget the chain rule

g'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2 chain rule again

g''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3 chain rule again

g'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4

g'''(x) = -48/[(2x -4)^4]
g'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16 = ***real ans***

g'''(3) = -3

Answer 2

Is it 1 / (2x-4)? if so, then write it (2x-4)^-1, then use the power rule to get -(2x-4)^-2 * 2 for the first derivative. Use the power rule again to get the second derivative 2*(2x-4)^-3 * 2 * 2; Do it once more to get the third derivative. Then plug in 3 for x, and yes, the answer is -3

d f(x)^n / dx = n*f(x)^(n-1) * d f(x) / dx

Answer 3

g'(x) = -2(2x-4)^-2

g''(x) = 8(2x-4)^-3

g'''(x) =-48(2x-4)^-4

= -48(2*3-4)^-4
= -48(2)^-4
= -48 * 16^-1
=-3