# The $43$rd derivative of $\sin(x^{13}+x^{15})$ | Calculus 1

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

• Taylor expand the sine. – hmakholm left over Monica Feb 3 '13 at 16:40
• 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 ... – hmakholm left over Monica Feb 3 '13 at 17:01
• yea I forgot writing it, thanks anyway! – ORBOT Inc. Feb 3 '13 at 17:19

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: $$-43!/2$$
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$$\sin(x^{13}+x^{15})=\sin x^{13}\cos x^{15}+\cos x^{13}\sin x^{15}$$
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.