Quadratic Reciprocal Higher Derivatives is a chronological lesson in higher-order differentiation. The source manuscript supplies this topic as part of Methods and Applications of Higher-Order Differentiation. :::note[Domain and Notation Convention] When expressions contain $x^2-a^2$, the decomposition using $2a$ requires $a\ne0$. For $x^2+a^2$, the phase angle must be chosen by both sine and cosine relations, not by a quadrant-ambiguous inverse tangent alone. ::: :::definition[Quadratic Reciprocal Form] Let $a\in\mathbb{R}$ and let the denominator be nonzero. A function of the form $(x^2-a^2)^{-1}$ or $(x^2+a^2)^{-1}$ is called a quadratic reciprocal form for this lesson. ::: :::theorem[Difference of Squares Reciprocal] Let $a\ne0$ and let $x\ne a,-a$. Then $$\begin{equation*} \frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right). \end{equation*}$$ Consequently, $$\begin{equation*} D^n\left(\frac{1}{x^2-a^2}\right)=\frac{(-1)^n n!}{2a}\left(\frac{1}{(x-a)^{n+1}}-\frac{1}{(x+a)^{n+1}}\right). \end{equation*}$$ ::: :::proof Given that $a\ne0$ and $x\ne a,-a$. To prove the displayed formula. Using partial fractions, $$\begin{align*} \frac{1}{x^2-a^2} &=\frac{1}{(x-a)(x+a)}\\ &=\frac{A}{x-a}+\frac{B}{x+a}. \end{align*}$$ Substituting $x=a$ gives $A=1/(2a)$, and substituting $x=-a$ gives $B=-1/(2a)$. Differentiating the two reciprocal terms $n$ times gives the stated result. Hence the formula holds on the stated domain. $\square$ ::: :::note[Sum of Squares] For $x^2+a^2$, use a branch-safe phase by writing $x=r\cos\theta$ and $a=r\sin\theta$, where $r=\sqrt{x^2+a^2}$. This fixes both sine and cosine and avoids quadrant ambiguity. ::: :::exercise[Practice Questions] 1. Decompose $1/(x^2-a^2)$ when $a\ne0$. 2. State the excluded values for $1/(x^2-a^2)$. 3. Why must the phase for $x^2+a^2$ be branch-safe? ::: :::answer[Answers and Guidance] 1. $1/(x^2-a^2)=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)$. 2. $x=a$ and $x=-a$ are excluded. 3. Because a single inverse tangent may not determine the correct quadrant; both sine and cosine relations determine the phase. ::: :::faq[Frequently Asked Questions] Q: What is the main purpose of this lesson? A: The purpose is to use quadratic reciprocal higher 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/cyclic-exponential-trigonometric-derivatives button-ghost: Review Higher Order Derivatives | /real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives :::