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

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

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

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

2. I've done the first derivative of the next question but i want to make sure its right before I move on.

ii) $\displaystyle 2x \ln(3x+1)$

Let $\displaystyle u = 2x$ and $\displaystyle v = \ln(3x+1)$

$\displaystyle \frac{du}{dx} = 2$ and $\displaystyle \frac{dv}{dx} = \frac{3}{3x+1}$

F of F

$\displaystyle 2 \ln(3x+1) + 2x(\frac{3}{3x+1})$

$\displaystyle = 2\ln(3x+1)+2x$

3. For the second derivative i believe there should be a -x where the -1 is in the numerator and therefore your answer would be -0.345588

4. The first derivative of the second one should be.

$\displaystyle 2 ln(3x+1) + \frac{6x}{3x+1}$

Use the chain rule and the rule for the derivative of ln(u) which is $\displaystyle \frac {u'}{u}$

5. If you want me to show you the steps just let me know

6. Originally Posted by mybrohshi5
For the second derivative i believe there should be a -x where the -1 is in the numerator and therefore your answer would be -0.345588
Can you see where I have gone wrong with this?

Originally Posted by mybrohshi5
The first derivative of the second one should be.

Use the chain rule and the rule for the derivative of ln(u) which is
I got that far but obviously tried to simplify it....incorrectly

Thanks for the input

7. 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.

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

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

8. Aaaah I see I missed a bit whilst copying my working down, thanks for that

9. No problem =) Good luck