Standard nth Derivatives is part of the study of higher-order differentiation. The lesson fixes the notation, states the required hypotheses, and then uses the supplied source material in chronological order. :::section[Source Development] The following development follows the uploaded Markdown source for this topic, with editorial correction of notation, domains, and branch conditions where needed. ::: ### Theorem: nth Derivative of a Real Power Let $k\in\mathbb{R}$ and let $n\in\mathbb{N}$. Suppose that the function $f(x)=x^k$ is $n$ times differentiable on the interval under consideration. Then $$\begin{equation*} \frac{d^n}{dx^n}(x^k) = k(k-1)(k-2)\cdots(k-n+1)x^{k-n}. \end{equation*}$$ Given that $f(x)=x^k$. To prove that $$\begin{equation*} D^n(x^k) = k(k-1)\cdots(k-n+1)x^{k-n}. \end{equation*}$$ We use mathematical induction on $n$. For $n=1$, $$\begin{equation*} D(x^k)=kx^{k-1}. \end{equation*}$$ Therefore, the result is true for $n=1$. Assume that the result is true for a positive integer $m$. Then $$\begin{equation} D^m(x^k) = k(k-1)\cdots(k-m+1)x^{k-m}. \end{equation}$$ Differentiating equation the displayed induction equation, $$\begin{align*} D^{m+1}(x^k) &= D\left[ k(k-1)\cdots(k-m+1)x^{k-m} \right]\\ &= k(k-1)\cdots(k-m+1)(k-m)x^{k-m-1}\\ &= k(k-1)\cdots\bigl(k-(m+1)+1\bigr)x^{k-(m+1)}. \end{align*}$$ Therefore, the result is true for $n=m+1$ whenever it is true for $n=m$. Hence, by mathematical induction, $$\begin{equation*} D^n(x^k) = k(k-1)\cdots(k-n+1)x^{k-n} \end{equation*}$$ for every positive integer $n$ for which the derivatives exist. $$\blacksquare$$ ### Corollary: Derivative of Order Equal to the Degree Let $k\in\mathbb{N}$ and let $y=x^k$. Then $$\begin{equation*} y^{(k)}=k!. \end{equation*}$$ Given that $y=x^k$ and $k\in\mathbb{N}$. Using the power-derivative theorem with $n=k$, $$\begin{align*} y^{(k)} &= k(k-1)(k-2)\cdots 2\cdot 1\,x^0\\ &=k!. \end{align*}$$ Hence, $$\begin{equation*} y^{(k)}=k!. \end{equation*}$$ $$\blacksquare$$ ### Corollary: Derivatives Beyond the Degree Let $k\in\mathbb{N}$ and let $n>k$. If $y=x^k$, then $$\begin{equation*} y^{(n)}=0. \end{equation*}$$ Given that $n>k$. The $(k+1)$th derivative contains the factor $$\begin{equation*} k-k=0. \end{equation*}$$ Therefore, $$\begin{equation*} D^{k+1}(x^k)=0. \end{equation*}$$ Every derivative of the zero function is zero. Hence, $$\begin{equation*} D^n(x^k)=0 \end{equation*}$$ for every $n>k$. $$\blacksquare$$ ### Corollary: nth Derivative of a Negative Power Let $k>0$ and let $x\neq 0$. Then $$\begin{equation*} \frac{d^n}{dx^n}(x^{-k}) = (-1)^n k(k+1)(k+2)\cdots(k+n-1)x^{-k-n}. \end{equation*}$$ Given that $y=x^{-k}$. Using the power-derivative theorem, $$\begin{align*} y^{(n)} &= (-k)(-k-1)(-k-2)\cdots(-k-n+1)x^{-k-n}\\ &= (-1)^n k(k+1)(k+2)\cdots(k+n-1)x^{-k-n}. \end{align*}$$ Hence, $$\begin{equation*} y^{(n)} = (-1)^n k(k+1)\cdots(k+n-1)x^{-k-n}. \end{equation*}$$ $$\blacksquare$$ ### Formula: Negative Integral Powers If $k\in\mathbb{N}$, then $$\begin{equation*} \frac{d^n}{dx^n}(x^{-k}) = (-1)^n \frac{(k+n-1)!}{(k-1)!x^{k+n}}. \end{equation*}$$ ### Formula: Reciprocal Function Taking $k=1$, $$\begin{equation*} \frac{d^n}{dx^n}\left(\frac{1}{x}\right) = \frac{(-1)^n n!}{x^{n+1}}. \end{equation*}$$ ### Formula: Reciprocal Square Taking $k=2$, $$\begin{equation*} \frac{d^n}{dx^n}\left(\frac{1}{x^2}\right) = \frac{(-1)^n(n+1)!}{x^{n+2}}. \end{equation*}$$ ### Theorem: nth Derivative of the Natural Logarithm Let $x>0$ and let $n\in\mathbb{N}$. Then $$\begin{equation*} \frac{d^n}{dx^n}(\log x) = \frac{(-1)^{n-1}(n-1)!}{x^n}. \end{equation*}$$ Given that $$\begin{equation*} \frac{d}{dx}(\log x)=\frac{1}{x}. \end{equation*}$$ The nth derivative of $\log x$ is the $(n-1)$th derivative of $x^{-1}$. Therefore, $$\begin{align*} D^n(\log x) &= D^{n-1}(x^{-1})\\ &= \frac{(-1)^{n-1}(n-1)!}{x^n}. \end{align*}$$ Hence, $$\begin{equation*} D^n(\log x) = \frac{(-1)^{n-1}(n-1)!}{x^n}. \end{equation*}$$ $$\blacksquare$$ ### Formula: Reciprocal of a Translated Variable For $x\neq a$, $$\begin{equation*} \frac{d^n}{dx^n}\left(\frac{1}{x-a}\right) = \frac{(-1)^n n!}{(x-a)^{n+1}}. \end{equation*}$$ ### Theorem: Reciprocal of a Linear Function Let $a,b\in\mathbb{R}$ with $a\neq 0$, and suppose that $ax+b\neq 0$. Then $$\begin{equation*} \frac{d^n}{dx^n}\left(\frac{1}{ax+b}\right) = \frac{(-1)^n n!a^n}{(ax+b)^{n+1}}. \end{equation*}$$ Given that $$\begin{equation*} y=(ax+b)^{-1}. \end{equation*}$$ For $n=1$, $$\begin{equation*} y' = -a(ax+b)^{-2} = \frac{-a}{(ax+b)^2}. \end{equation*}$$ Therefore, the result is true for $n=1$. Assume that $$\begin{equation*} y^{(m)} = \frac{(-1)^m m!a^m}{(ax+b)^{m+1}}. \end{equation*}$$ Differentiating, $$\begin{align*} y^{(m+1)} &= (-1)^m m!a^m D\left((ax+b)^{-m-1}\right)\\ &= (-1)^m m!a^m(-m-1)a(ax+b)^{-m-2}\\ &= \frac{(-1)^{m+1}(m+1)!a^{m+1}} {(ax+b)^{m+2}}. \end{align*}$$ Therefore, the result is true for $m+1$. Hence, by mathematical induction, $$\begin{equation*} \frac{d^n}{dx^n}\left(\frac{1}{ax+b}\right) = \frac{(-1)^n n!a^n}{(ax+b)^{n+1}}. \end{equation*}$$ $$\blacksquare$$ ### Exercise Question: Verification of a Reciprocal Formula Prove by mathematical induction that $$\begin{equation*} D^n\left(\frac{1}{ax+b}\right) = \frac{(-1)^n n!a^n}{(ax+b)^{n+1}}. \end{equation*}$$ ### Answer The required result is the preceding theorem. --- :::exercise[Practice Questions] 1. Find $\dfrac{d^n}{dx^n}x^m$ for integers $m\geq n\geq0$. 2. Find the $n$th derivative of $1/(x-a)$ for $x\ne a$. 3. Find the $n$th derivative of $\log x$ on $(0,\infty)$ for $n\geq1$. ::: :::answer[Answers and Guidance] 1. $\dfrac{m!}{(m-n)!}x^{m-n}$. 2. $(-1)^n n!/(x-a)^{n+1}$. 3. $(-1)^{n-1}(n-1)!/x^n$. ::: :::faq[Frequently Asked Questions] Q: What is the main purpose of this lesson? A: The purpose is to use standard nth derivatives in a controlled way, with all variables, orders, and restrictions stated before calculation. Q: What is the most common error in this topic? A: The most common error is to apply a formula without checking the differentiability, domain, or nonzero-denominator condition required by the formula. Q: How should I verify an answer? A: Differentiate the obtained expression in a low-order case, compare it with the stated formula, and confirm that the excluded values have not been lost. ::: :::review-history[Academic Review Notes] - Source mathematics reviewed against the uploaded higher-order differentiation manuscript. - Domains, branch conventions, exercises, answers, FAQ formatting, and final call-to-action placement checked. ::: :::call-to-action[Continue Learning] subtitle: Revise the formulae, complete the practice questions, and continue to the next lesson in higher-order differentiation. button: Next Lesson | /real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives/quadratic-reciprocal-higher-derivatives button-ghost: Review Higher Order Derivatives | /real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives :::