Originally Posted by

**anothernewbie** Hi would somebody be able to check my answers to some questions I have, i'll just post one at a time as they take up a bit of room:

Q: Obtain $\displaystyle f''(0.3)$ for each of the following functions.

i) $\displaystyle f(x) = \arccos x$

A:

$\displaystyle f(x) = \arccos x$

$\displaystyle f'(x) = -(1-x^2)^ \frac{-1}{2}$

Sub $\displaystyle z = 1-x^2$

$\displaystyle y = -z^\frac{-1}{2}$

$\displaystyle \frac{dy}{dz} = \frac{1}{2}(1-x^2)^\frac{-3}{2}$

right here is where you went wrong. you need to apply the chain rule so you need to include this $\displaystyle \frac{dy}{dz} = \frac{1}{2}(1-x^2)^\frac{-3}{2}$ multiplied by a -2x because the -2x is the derivative of 1-x^2

The 2's cancel out and you are left with -x on the top.

$\displaystyle f''(x) = \frac{-1}{\sqrt{(1-x^2)^3}}$

$\displaystyle f''(0.3) = 1.152$