1. ## Second Derivative Problem

given that $h(x) = \frac{1}{(3x-5)^2^'}$.Find $h''(1)$

2. Have you tried the chain rule? what did you get?

3. i have no idea sir.I confused with the 2 prime sign.

4. "double prime" just means differentiate twice, ie obtain the derivative, then differentiate the derivative.

5. i mean the prime at the fraction.

6. i assumed that was a typo! It has no standard meaning that i'm aware of.

If it has a meaning that is particular to your course/teacher then the answer is probably in your course notes or textbook.

7. ok..when i using chain rule.I get.
$h(x) = (3x-5)^{-2}$

$\frac{dy}{dx} = 3(-2) (3x-5) ^ {-2-1} \times \frac{d}{dx} (3x-5)$

$=-6(3x-5)^{-3} \times 3$

$= \frac{-18}{(3x-5)^3}$

is that correct?then what should i do?

8. that isn't correct. You pulled out a factor of "3" on your second line that isn't needed. h'(x) should be:

$h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-2-1) \times \dfrac{d}{dx}(3x-5)$

$h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-3) \times 3$

$h'(x) = \dfrac{dy}{dx} = -6(3x-5)^{-3)$

$h'(x) = \dfrac{dy}{dx} = \frac{-6}{(3x-5)^3}$

Differentiate again to get h''(x). Then evaluate your answer at x=1 to find h''(1).