Given the following values of x, g(x) and g'(x), what is h'(2/3) if h(x)=g(2/x)?

[BifSlamkovich]

1. Homework Statement
x g(x) g'(x)
-2 -2 2
-1 0 1
0 1 2
1 3 4
2 7 3
3 9 2


2. Homework Equations

chain rule

3. The Attempt at a Solution

differentiate (2/x) and multiply that times g'(2/x). Plug in 2/3 into -2/(x^2), and one obtains -9/2. g'(2/((2/3)))= 2. The two multiplied together yields -9. Is that correct?

 

[SammyS]

Please, Don't double-post questions.

Where's the problem statement? ... Oh! It's there in the title line. (It's a good idea to repeat the question in the main text.)

Yes, d/dx (g(2/x)) evaluated at x = 2/3 is g'(2/(2/3))∙(-2/(2/3)²).

 

[BifSlamkovich]

Please, Don't double-post questions.

Where's the problem statement? ... Oh! It's there in the title line. (It's a good idea to repeat the question in the main text.)

Yes, d/dx (g(2/x)) evaluated at x = 2/3 is g'(2/(2/3))∙(-2/(2/3)²).

Sorry about not posting the problem statement in the main text. And I double-posted because it was first in the wrong category. This is calc., not pre-calc.