• Dec 9th 2009, 12:16 PM
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 $\displaystyle z = 1-x^2$
$\displaystyle y = -z^\frac{-1}{2}$

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

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

$\displaystyle f''(0.3) = 1.152$
• Dec 9th 2009, 12:24 PM
anothernewbie
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

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

$\displaystyle = 2\ln(3x+1)+2x$
• Dec 9th 2009, 12:26 PM
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
• Dec 9th 2009, 12:31 PM
mybrohshi5
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 $\displaystyle \frac {u'}{u}$
• Dec 9th 2009, 12:33 PM
mybrohshi5
If you want me to show you the steps just let me know (Nod)
• Dec 9th 2009, 12:37 PM
anothernewbie
Quote:

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?

Quote:

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

http://www.mathhelpforum.com/math-he...8e6085d4-1.gif

Use the chain rule and the rule for the derivative of ln(u) which is http://www.mathhelpforum.com/math-he...0e7cfe50-1.gif

I got that far but obviously tried to simplify it....incorrectly

Thanks for the input
• Dec 9th 2009, 12:49 PM
mybrohshi5
Quote:

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
• Dec 9th 2009, 12:56 PM
anothernewbie
Aaaah I see I missed a bit whilst copying my working down, thanks for that
• Dec 9th 2009, 01:01 PM
mybrohshi5
No problem =) Good luck