given that $\displaystyle h(x) = \frac{1}{(3x-5)^2^'}$.Find $\displaystyle h''(1)$
ok..when i using chain rule.I get.
$\displaystyle h(x) = (3x-5)^{-2}$
$\displaystyle \frac{dy}{dx} = 3(-2) (3x-5) ^ {-2-1} \times \frac{d}{dx} (3x-5)$
$\displaystyle =-6(3x-5)^{-3} \times 3$
$\displaystyle = \frac{-18}{(3x-5)^3}$
is that correct?then what should i do?
that isn't correct. You pulled out a factor of "3" on your second line that isn't needed. h'(x) should be:
$\displaystyle h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-2-1) \times \dfrac{d}{dx}(3x-5)$
$\displaystyle h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-3) \times 3$
$\displaystyle h'(x) = \dfrac{dy}{dx} = -6(3x-5)^{-3)$
$\displaystyle h'(x) = \dfrac{dy}{dx} = \frac{-6}{(3x-5)^3}$
Differentiate again to get h''(x). Then evaluate your answer at x=1 to find h''(1).