Answer check please

**anothernewbie** · December 9th 2009, 12:16 PM

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 $f''(0.3)$ for each of the following functions.

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

A:

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

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

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

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

$f''(0.3) = 1.152$

**anothernewbie** · December 9th 2009, 12:24 PM

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

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

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

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

F of F

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

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

**mybrohshi5** · December 9th 2009, 12:26 PM

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

**mybrohshi5** · December 9th 2009, 12:31 PM

The first derivative of the second one should be.

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

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

**mybrohshi5** · December 9th 2009, 12:33 PM

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

**anothernewbie** · December 9th 2009, 12:37 PM

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

**mybrohshi5** · December 9th 2009, 12:49 PM

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 $f''(0.3)$ for each of the following functions.

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

A:

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

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

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

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

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

$f''(0.3) = 1.152$

See above

**anothernewbie** · December 9th 2009, 12:56 PM

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

**mybrohshi5** · December 9th 2009, 01:01 PM

No problem =) Good luck

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