# 10th derivative

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• Aug 20th 2009, 08:21 AM
dat1611
10th derivative
Compute the 10th derivative of

http://webwork.umemat.maine.edu/webw...8af4ef5f81.png

at x=0
f^10(0)=?

i think im supposed to use the Maclaurin series
• Aug 20th 2009, 09:39 AM
CaptainBlack
Quote:

Originally Posted by dat1611
Compute the 10th derivative of

http://webwork.umemat.maine.edu/webw...8af4ef5f81.png

at x=0
f^10(0)=?

i think im supposed to use the Maclaurin series

Use the McLaurin series for $\displaystyle \arctan(u)$ and substitute $\displaystyle u=x^2/4$.

CB
• Aug 20th 2009, 10:02 AM
dat1611
i dont know how im supposed to get a number answer out of it
• Aug 20th 2009, 01:24 PM
CaptainBlack
Quote:

Originally Posted by dat1611
i dont know how im supposed to get a number answer out of it

You get the McLaurin series up to the term in $\displaystyle x^{10}$ then you can differentiate term by term ten times (which is easy as the terms with exponent of $\displaystyle x$ less than $\displaystyle 10$ will be eliminated by the time you have taken the 10-th derivative) and then set $\displaystyle x=0$ to evaluate (the terms with powers greater than $\displaystyle 10$ when $\displaystyle x$ is set to zero will be zero so there was no need to look at them).

CB
• Aug 20th 2009, 04:37 PM
dat1611
Im not sure how to do the Maclaurin series
• Aug 20th 2009, 08:05 PM
CaptainBlack
Quote:

Originally Posted by dat1611
Im not sure how to do the Maclaurin series

You should know the series:

$\displaystyle \arctan(u)=u-{{u^3}\over{3}}+{{u^5}\over{5}}-{{u^7}\over{7}}+ {{u^9}\over{9}}..$

you will not need the terms beyond $\displaystyle u^5$

CB
• Aug 21st 2009, 07:45 AM
dat1611
how do i get a number if i plug in x^2/4
• Aug 21st 2009, 07:53 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
You should know the series:

$\displaystyle \arctan(u)=u-{{u^3}\over{3}}+{{u^5}\over{5}}-{{u^7}\over{7}}+ {{u^9}\over{9}}..$

you will not need the terms beyond $\displaystyle u^5$

CB

Quote:

Originally Posted by dat1611
how do i get a number if i plug in x^2/4

Substitute $\displaystyle x^2/4$ for $\displaystyle u$ to get:

$\displaystyle \arctan(x^2/4)=\frac{x^2}{4}-{{x^6}\over{3.4^3}}+{{x^{10}}\over{5.4^5}}-{{x^{14}}\over{7.4^7}} + ..$

truncate after the term with $\displaystyle x^{10}$ as you won't need them (Why is that?) and diffterentiate 10 times (you do that), then substitute $\displaystyle 0$ for $\displaystyle x$ in what you have left (if anything).

CB
• Aug 21st 2009, 08:04 AM
dat1611
so i differntiated x^2/4- x^6/12 + x^10/20 ten times and all i got is 181440 which is not right
• Aug 21st 2009, 09:42 AM
Random Variable
$\displaystyle \arctan(u) = u - \frac{u^{3}}{3} + \frac{u^{5}}{5} - \frac{u^{7}}{7} + ...$

$\displaystyle \arctan (x^{2}/4) = \frac{x^{2}}{4} - \frac{x^{6}}{3\cdot4^3} + \frac{x^{10}}{5 \cdot 4^{5}} + \frac{x^{14}}{7 \cdot4^{7}} + ...$

$\displaystyle \frac{d^{n}}{dx^{n}} x^{k} = \frac{k!}{(k-n)!} x^{k-n}$

The 10th derivatives of the first two terms are zero for any value of x, and the 10th derivative of every term beyond $\displaystyle x^{10}$ evaluates to zero at x=0.

so $\displaystyle \frac{d^{10}}{dx^{10}} \arctan \Big(\frac{x^{2}}{4}\Big) = \frac{1}{5 \cdot 4^5} \frac{d^{10}}{dx^{10}} x^{10} = \frac{1}{5 \cdot 4^{5}}\frac{10!}{0!} x^{0} = \frac{2835}{4}$

which of course at $\displaystyle x=0$ is $\displaystyle \ \frac{2835}{4}$