## Nth Derivative of a Product :::theorem[Leibniz Formula] Let $u$ and $v$ be functions of $x$ that are differentiable at least $n$ times. Then $$\begin{equation*} D^n(uv)=\sum_{r=0}^{n}\binom{n}{r}u^{(n-r)}v^{(r)}. \end{equation*}$$ ::: :::proof Given that $u$ and $v$ possess derivatives through order $n$. To prove that the stated formula holds. For $n=1$, the formula is the ordinary product rule. Suppose it holds for a fixed $n$. Differentiating, $$\begin{align*} D^{n+1}(uv) &=\sum_{r=0}^{n}\binom{n}{r} \left(u^{(n-r+1)}v^{(r)}+u^{(n-r)}v^{(r+1)}\right)\\ &=\sum_{r=0}^{n+1}\binom{n+1}{r}u^{(n+1-r)}v^{(r)}, \end{align*}$$ using Pascal's identity. Hence the formula holds for $n+1$, and therefore for every positive integer $n$. $\square$ ::: :::example[Polynomial and Logarithm] Let $y=x^3\log x$ for $x>0$. Since derivatives of $x^3$ vanish after order $3$, Leibniz's formula contains at most four nonzero terms. For $n\geq4$, $$\begin{equation*} y_n= x^3D^n(\log x)+3nx^2D^{n-1}(\log x) +3n(n-1)xD^{n-2}(\log x) +n(n-1)(n-2)D^{n-3}(\log x). \end{equation*}$$ ::: ## Differential Recurrence Relations :::theorem[Repeated Differentiation of a Differential Equation] If $$\begin{equation*} (1-x^2)y_2-xy_1+m^2y=0, \end{equation*}$$ then repeated differentiation gives $$\begin{equation*} (1-x^2)y_{n+2}-(2n+1)xy_{n+1}+(m^2-n^2)y_n=0. \end{equation*}$$ ::: At $x=0$, $$\begin{equation*} (y_{n+2})_0=-(m^2-n^2)(y_n)_0. \end{equation*}$$ :::section[Method and Verification] Write the Leibniz sum with a consistent index. Identify derivatives that vanish, simplify only after the complete sum is written, and verify the result for small values of $n$. For recurrence relations, differentiate the defining identity before substituting indexed notation. ::: :::exercise[Practice Questions] 1. Use Leibniz's formula to find $D^n(x^2e^{ax})$. 2. Find $D^n(x^3\log x)$ for $n\geq4$. 3. State which terms vanish in Leibniz's formula when $u$ is a polynomial of degree $m$. ::: :::answer[Answers and Solution Guidance] 1. $D^n(x^2e^{ax})=e^{ax}[a^nx^2+2na^{n-1}x+n(n-1)a^{n-2}]$. 2. Substitute $D^r(\log x)=(-1)^{r-1}(r-1)!/x^r$ into the four surviving Leibniz terms. 3. Since $u^{(r)}=0$ for $r>m$, every term containing such a derivative vanishes. ::: :::faq[Frequently Asked Questions] Q: Why do binomial coefficients appear in Leibniz's formula? A: They count the ways in which the $n$ differentiations can be distributed between the two factors. Q: When does the Leibniz sum terminate early? A: It terminates when one factor is a polynomial, because its derivatives above the polynomial degree are zero. Q: How should a recurrence obtained by repeated differentiation be checked? A: Verify the original case and one differentiated case, then compare coefficients and derivative orders carefully. ::: :::call-to-action[Continue Learning] subtitle: Complete the practice and continue to the advanced repeated-differentiation problems. button: Next Lesson|/real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives/advanced-problems-in-repeated-differentiation button-ghost: Review Higher-Order Derivatives|/real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives :::