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
 A: 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:
$$-43!/2$$
A: 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.
