# Math Help - 10th derivative

1. ## 10th derivative

Compute the 10th derivative of

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

i think im supposed to use the Maclaurin series

2. Originally Posted by dat1611
Compute the 10th derivative of

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

i think im supposed to use the Maclaurin series
Use the McLaurin series for $\arctan(u)$ and substitute $u=x^2/4$.

CB

3. i dont know how im supposed to get a number answer out of it

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

CB

5. Im not sure how to do the Maclaurin series

6. Originally Posted by dat1611
Im not sure how to do the Maclaurin series
You should know the series:

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

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

CB

7. how do i get a number if i plug in x^2/4

8. Originally Posted by CaptainBlack
You should know the series:

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

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

CB
Originally Posted by dat1611
how do i get a number if i plug in x^2/4
Substitute $x^2/4$ for $u$ to get:

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

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

CB

9. so i differntiated x^2/4- x^6/12 + x^10/20 ten times and all i got is 181440 which is not right

10. $\arctan(u) = u - \frac{u^{3}}{3} + \frac{u^{5}}{5} - \frac{u^{7}}{7} + ...$

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

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

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

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

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