Leibniz's formula

Before we discuss Leibniz's formula, let us go through an example where we compute derivatives of order higher than 1. Consider the function $f(x)=\ln x$. The first four derivatives are given by

From the first few derivatives we can observe that the $n^{\text {th }}$ derivative is given by,

Next, we introduce the Leibniz derivative rule which allows us to express the $n^{\text {th }}$ derivative of a product (i.e. it is a generalisation of the product rule to higher derivatives). Consider two functions of $x, u(x)$ and $v(x)$; the derivative of their product:

Differentiating again with respect to $x$ gives

and a third time,

It may be clear at this point that the numerical coefficients are the same as the ones in the binomial theorem:

Now, for $n=2$,

note that we included $a^{(0)}$ and $b^{(0)}$ so that we can make a direct comparison to $f^{(2)}$. Recall that the zeroth derivative of a function is simply the function itself e.g. so $u^{(0)}=u(x)$ and $v^{(0)}=v(x)$. Similarly, for $n=3$ and $m=0, \cdots, 3$, we obtain the coefficients 1, 3, 3, 1 as they appear in $f^{(3)}$. The Leibniz rule (Note that this is different to the Leibniz integral rule which is used for differentiation under the integral sign) is given by the following formula:

where the binomial coefficients are given by

The formula is particularly useful when either $v$ (or $u$ ) are polynomials so that the derivatives $v^{(m)}$ are zero apart from the first few terms. For example, if we wanted to differentiate $n$ times the function $f(x)=x^{2} e^{2 x}$, we express the function $f$ as the product of two functions, $u(x)=e^{2 x}$ and $v(x)=x^{2}$. Then, since $v^{\prime}=2 x, v^{\prime \prime}=2$, and $v^{\prime \prime \prime}=0$ for $m=1,2$, and 3, respectively, $v^{(m)}=0$ for $m \geq 3$. Application of the rule gives,

We note that $u^{(m)}=2^{m} e^{2 x}$ and so

using