I know how to do it with easier functions, but is there a universal method which can be applied to all continuous functions differentiable at least $n$ times(introduced to in a second year calculus class)?
I can do it for easy ones like $\sin(x)$ and $\cos(x)$, but $\sin(x^2)$ and $\tan(x)$ and $\ln(1+x^2)$ are proving to be very difficult.
Thanks.
There are four parts in this answer.
The Faa di Bruno formula (see Theorem 11.4 in the book [1] and Theorem C on page 139 in the monograph [2] below) can be described in terms of the Bell polynomials of the second kind $B_{n,k}\bigl(x_1,x_2,\dotsc,x_{n-k+1}\bigr)$ by \begin{equation}\label{Bruno-Bell-Polynomial}\tag{1} \frac{\textrm{d}^n}{\textrm{d}x^n}f\circ h(x)=\sum_{k=0}^nf^{(k)}(h(x)) B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr), \quad n\ge0. \end{equation} In Theorem 5.1 of the paper [3] and Section 3 of the paper [4] below, it was found that \begin{equation}\label{Bell-x-1-0-eq}\tag{2} B_{n,k}(x,1,0,\dotsc,0) =\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}, \quad n\ge k\ge0, \end{equation} where $\binom{0}{0}=1$ and $\binom{p}{q}=0$ for $q>p\ge0$. Consequently, as an example, we have \begin{equation*} \Bigl(e^{x^2}\Bigr)^{(n)} =e^{x^2}\frac{n!}{(2x)^n} \sum_{k=0}^{n}\binom{k}{n-k}\frac{(2x)^{2k}}{k!}, \quad n\ge0. \end{equation*} Similarly, by virtue of the formulas \eqref{Bruno-Bell-Polynomial} and \eqref{Bell-x-1-0-eq}, general formulas for the $n$th derivatives of the functions $\sin\bigl(x^2\bigr)$ and $\ln\bigl(1+x^2\bigr)$ can be derived readily as \begin{equation}\tag{+} \bigl[\sin\bigl(x^2\bigr)\bigr]^{(n)} =\frac{n!}{(2x)^n}\sum_{k=1}^n\binom{k}{n-k}\frac{(2x)^{2k}}{k!} \sin\biggl(x^2+\frac{k\pi}{2}\biggr), \quad n\ge0 \end{equation} and \begin{equation}\tag{#} \bigl[\ln\bigl(1+x^2\bigr)\bigr]^{(n)} =\frac{n!}{(2x)^n}\sum_{k=1}^n\binom{k}{n-k}\frac{(-1)^{k-1}}{k}\frac{(2x)^{2k}}{(1+x^2)^{k}}, \quad n\ge1. \end{equation}
In Theorem 1.2 of the paper [5] below, it was derived that \begin{multline}\label{bell-sin-eq}\tag{3} B_{n,k}\biggl(-\sin x,-\cos x,\sin x,\cos x,\dotsc, \cos\biggl[x+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\ =\frac{(-1)^k\cos^kx}{k!}\sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^\ell}{(2\cos x)^\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)x+\frac{n\pi}2\biggr] \end{multline} and \begin{multline}\label{bell-sin=ans}\tag{4} B_{n,k}\biggl(\cos x,-\sin x,-\cos x,\sin x,\dotsc, \sin\biggl[x+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\ =\frac{(-1)^k\sin^{k}x}{k!}\sum_{\ell=0}^k\binom{k}{\ell}\frac1{(2\sin x)^{\ell}} \sum_{q=0}^\ell(-1)^q\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)x+\frac{(n-\ell)\pi}2\biggr] \end{multline} for $n\ge k\ge0$. Since $\sin\bigl(x\pm\frac{\pi}{2}\bigr)=\pm\cos x$ and $\cos\bigl(x\pm\frac{\pi}{2}\bigr)=\mp\sin x$, the formulas \eqref{bell-sin-eq} and \eqref{bell-sin=ans} are equivalent to each other. These two formulas \eqref{bell-sin-eq} and \eqref{bell-sin=ans} can be applied to establish general formulas of the $n$th derivatives for functions of the types $f(\sin x)$ and $f(\cos x)$, such as $\sin^\alpha x$, $\cos^\alpha x$, $\sec^\alpha x$, $\csc^\alpha x$, $e^{\pm\sin x}$, $e^{\pm\cos x}$, $\ln\cos x$, $\ln\sin x$, $\ln\sec x$, $\ln\csc x$, $\sin\sin x$, $\cos\sin x$, $\sin\cos x$, $\cos\cos x$, $\tan x=-(\ln\cos x)'$, and $\cot x=(\ln \sin x)'$, if the general formula for the $n$th derivative of $f$ is computable.
Define the falling factorial of $\alpha\in\mathbb{C}$ by \begin{equation*}%\label{Fall-Factorial-Dfn-Eq} \langle\alpha\rangle_n= \prod_{k=0}^{n-1}(\alpha-k)= \begin{cases} \alpha(\alpha-1)\dotsm(\alpha-n+1), & n\ge1;\\ 1,& n=0. \end{cases} \end{equation*} In Remark 3.1 on page 88 in the paper [17] below, the formula \begin{equation}\label{Bell-1-lambda}\tag{5} B_{n,k}\Biggl(1, 1-\lambda, (1-\lambda)(1-2\lambda),\dotsc, \prod_{\ell=0}^{n-k}(1-\ell\lambda)\Biggr) =\frac{(-1)^k}{k!} \sum_{\ell=0}^k (-1)^{\ell} \binom{k}{\ell} \prod_{q=0}^{n-1}(\ell-q\lambda) \end{equation} for $\lambda\in\mathbb{C}$ and $n\ge k\ge0$ was concluded. In Theorem 2.1 on page 165 in the paper [11] below, it was discovered that \begin{equation}\label{Bell-fall-Eq}\tag{6} B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1}) =\frac{(-1)^k}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\langle\alpha\ell\rangle_n \end{equation} for $\alpha\in\mathbb{C}$ and $n\ge k\ge0$. These two formulas \eqref{Bell-1-lambda} and \eqref{Bell-fall-Eq} are equivalent to each other.
By virtue of the Faa di Bruno formula \eqref{Bruno-Bell-Polynomial} and any one of the formulas \eqref{Bell-1-lambda} and \eqref{Bell-fall-Eq}, one can establish general formulas of the $n$th derivatives for functions of the type $f(x^\alpha)$ for $\alpha\in\mathbb{R}$, such as $e^{x^\alpha}$, $\sin[(a+bx)^\alpha]$, and $\ln(1\pm x^\alpha)$, if the $n$th derivative of the function $f$ is computable.
The $n$th derivative formulas for the tangent function $\tan x=\dfrac{\sin x}{\cos x}$ and the cotangent function $\cot x=\dfrac{\cos x}{\sin x}$ can also be computed by the formula \begin{equation}\label{Sitnik-Bourbaki}\tag{7} \frac{\textrm{d}^k}{\textrm{d}z^k}\biggl(\frac{u}{v}\biggr) =\frac{(-1)^k}{v^{k+1}} \begin{vmatrix} u & v & 0 & \dotsm & 0\\ u' & v' & v & \dotsm & 0\\ u'' & v'' & 2v' & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \ddots & \dotsm\\ u^{(k-1)} & v^{(k-1)} & \binom{k-1}1v^{(k-2)} & \dots & v\\ u^{(k)} & v^{(k)} & \binom{k}1v^{(k-1)} & \dots & \binom{k}{k-1}v' \end{vmatrix}. \end{equation} For details on the formula \eqref{Sitnik-Bourbaki}, please refer to https://math.stackexchange.com/a/4261705/945479.
The texts in the first three parts above are extracted and modified from Sections 1.3, 1.4, and 1.6 in the paper [6] below.
References
The functions that you list, $\sin(x^2)$ and $\tan(x)$ and $\ln(1+x^2)$ are indeed infinitely many times differentiable, you just need to learn differentiation rules, e.g. product, quotient, chain, power rule, etc, and differentiate these functions as many times as you wish (and, perhaps, in some cases find a general formula for the $n$th derivative). Well, for some functions it will indeed be difficult to find a general formula. You may take the first few derivatives, and try to guess the formula for the $n$th derivative in each specific case. If you guess is correct, then you may prove it by induction.
No there is no universal method for differentiating functions, if you want to get other functions out. If you reduce it to a Taylor series, then you can do that, but the output of the process is another Taylor series.
Find the Taylor series, and use that to evaluate at the nth value?
Consider e.g. the Elementary functions, the Liouvillian functions, some of the Special functions and/or algebraic operations.
The set of the function terms of the Elementary functions is closed regarding differentiation.
The set of the function terms of the Liouvillian functions is closed regarding differentiation.
Let $m,n\in\mathbb{N}_0$ denote variables.
Some functions have an $n$-th derivative that is equal to its $m$-th derivative.
There are rules for the $n$-th derivative of some elementary standard functions. See e.g. Bronshtein/Semendyayev/Musiol/Mühlig: Handbook of Mathematics, Springer, table 6.3.
As for the first derivative, there are higher sum rule, higher factor rule, higher product rule (General Leibniz rule) and higher chain rule (Faà di Bruno's formula) to calculate the $n$-th derivatives.
You have to build your given function term hierarchically from smaller function terms and algebraic operations (unary or multiary algebraic functions) to apply the higher differentiation rules above.
$sin(x^2)$ and $ln(1+x^2)$ can be represented as compositions.
$tan(x)=\frac{sin(x)}{cos(x)}$ can be represented as quotient.
Apply higher chain rule and higher quotient rule respectively and hope that the smaller function terms have known general terms for their n-th derivatives.
In 2012, I wrote a still unpublished article "On partial Bell polynomials for the higher derivatives of composed functions".
