Question

find the derivative of
g(x)=2x+1/3x-1

Answer 1

For the sake of clarity, please use parentheses in expressions such as 1/3x. It's hard to know whether you mean (1/3)*x or 1/(3x).

Assuming that you mean (1/3)*x, the first power of x, with a constant, fractional coefficient, just differentiate each of the three terms of your expression separately:

g'(x) = 2 + (1/3) = 7/3.

Assuming that you mean 1/(3x), the reciprocal of 3x, rewrite g(x) as

g(x) = 2x + (3x)^(-1) - 1 first. Apply the power and constant coefficient rules to each term as appropriate.

Result: g'(x) = 2 -(3x)^(-2) (considerably different from before, no?)

Answer 2

oh ok i meant..\[g(x)=\frac{ 2x+1 }{ 3x-1 }\]

Answer 3

Thanks. Your 2nd expression is so much clearer it's like night and day. Are you OK with finding the derivative of g(x) (using the quotient rule)?

Answer 4

pretty much i just am having trouble figuring out why the answer has a negative sign?

Answer 5

I agree that the derivative of g(x) has a negative sign.
Have you written out all of the steps of the quotient rule?
(3x-1)(2) - (2x+1)(3)
g'(x) = ------------------
(3x-1)^2

Try simplifying/evaluating this and see if you don't get a negative result also.

Answer 6

oh i did get a negative now, my bad i was doing (2x+1)(3)-(3x-1)(2)

Answer 7

So you're OK now? Further questions?