Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A question from a test of my professor on calculus 1: Find the $43$rd derivative of $\sin(x^{13}+x^{15})$ at $x=0$.

Any idea I had didn't work, BTW, good luck to me on the test next week :X

share|improve this question
5  
Taylor expand the sine. –  Henning Makholm Feb 3 '13 at 16:40
2  
Tried it, I got this 3 degree polynom: x^13+x^15-x^41/2-x^43/2-x^45/6 how can I use it formalic? does that mean the answer is -43!/2 ? –  ORBOT Inc. Feb 3 '13 at 16:48
    
@ORBOTInc.: Yes, something like that, but I haven't checked the details. There should be an $x^{39}$ term somewhere ... –  Henning Makholm Feb 3 '13 at 17:01
    
yea I forgot writing it, thanks anyway! –  ORBOT Inc. Feb 3 '13 at 17:19

2 Answers 2

Expand $\sin(x)$ around zero to get: $$\sin(x)\sim x-\frac{x^3}{6}$$ $$\sin(x^{13}+x^{15})\sim x^{13}+x^{15}-\frac{x^{39}}{6}-\frac{x^{41}}{2}-\frac{x^{43}}{2}-\frac{x^{45}}{6}+O\left(x^{46}\right)$$ Take the $43$'rd derivative to get: $$-43!/2$$

share|improve this answer

First of all Use the trig indentity $\sin (A+B)=\sin A\cos B+\cos A\sin B$ to expand $\sin(x^{13}+x^{15})$.Here it is$$\sin(x^{13}+x^{15})=\sin x^{13}\cos x^{15}+\cos x^{13}\sin x^{15}$$

A little bit of theory to get started.

You can use Leibnitz's theorem for $nth$ derivative of a product of two functions.If $y=u v$,where $u$ and $v$ are functions of $x$,then$$y\prime=u v\prime+v u\prime$$where $v\prime=\frac{\mathrm{dv} }{\mathrm{d} x}$ and $u\prime=\frac{\mathrm{du} }{\mathrm{d} x}$ and $y\prime\prime=u v\prime\prime+v\prime u\prime+v u\prime\prime+u\prime v\prime=u\prime\prime v+2 u\prime v\prime+u v\prime\prime$.I would encourage you to get crazy and differentiate this (with a slight change in notation from $u\prime$ to $u^{(4)}$(fourth derivative)) as much as you want till you start seeing a pattern in which you will notice that in each case the superscript of $u$ decreases regularly by 1 and the superscript of $v$ increases regularly by 1 and the numerical coefficients are the normal binomial coefficients.

With that being said the $nth$ derivative can be computed using the binomial theorem.I mean $(u v)^{(n)}$ can be obtained by expanding $(u + v)^{(n)}$ which leads to Leibnitz's theorem$$y^{(n)}=\sum_{r=0}^{n}\binom{n}{r}u^{(n-r)}v^{(r)}$$

Now use the above formula to compute the $43rd$ derivative seperataly and see what you get.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.