Leibnitz's Formula for the $n$-th derivative of a product

This formula expresses the $n$-th derivative of the product of two variables in terms of the variables themselves and their successive derivatives.

If u and v are functions of $x$, we have, from equation (V) in §5.1 above,

$$\frac{d}{dx} (uv) = \frac{du}{dx} v + u \frac{dv}{dx}.$$

Differentiating again with respect to $x$,

$$\frac{d^2}{dx^2}(uv) = \frac{d^2u}{dx^2}v + \frac{du}{dx} \frac{... ...= \frac{d^2 u}{dx^2}v + 2 \frac{du}{dx} \frac{dv}{dx} + u \frac{d^2 v}{dx^2}.$$

Similarly,

$$\begin{array}{ll} \frac{d^3}{dx^3}(uv) &= \frac{d^3 u}{dx^3... ...{du}{dx} \frac{d^2 v}{dx^2} + u \frac{d^3 v}{dx^3}. \end{array}$$

However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond^7.1 to the exponents of the Binomial Theorem. Reasoning then by mathematical induction from the $m$-th to the $(m + 1 )$-st derivative of the product, we can prove Leibnitz's Formula

$$\frac{d^n}{dx^n}(uv) = \frac{d^n u}{dx^n} v + n \frac{d^{n - 1} u... ... \cdots + n \frac{du}{dx} \frac{d^{n - 1} v}{dx^{n - 1}} + u \frac{d^nv}{dx^n},$$ (7.1)

Example 7.4.1   Given $y = e^x\log\, x$, find $\frac{d^3 y}{dx^3}$ by Leibnitz's Formula.

Solution. Let $u = e^x$, and $v = \log\, x$; then $\frac{du}{dx} = e^x$, $\frac{dv}{dx} = \frac{1}{x}$, $\frac{d^2 u}{dx^2} = e^x$, $\frac{d^2 v}{dx^2} = - \frac{1}{x^2}$, $\frac{d^3 u}{dx^3} = e^x$, $\frac{d^3 v}{dx^3} = \frac{2}{x^3}$.

Substituting in (7.1), we get

$$\frac{d^3 y}{dx^3} = e^x \log x + \frac{3 e^x}{x} - \frac{3 e^x}{x^2} = e^x \left ( \log x + \frac{3}{x} - \frac{3}{x^2} + \frac{2}{x^3} \right ).$$

This can be verified using the SAGE commands:

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: f = exp(x)*log(x)
sage: diff(f,x,3)
e^x*log(x) + 3*e^x/x - 3*e^x/x^2 + 2*e^x/x^3

Example 7.4.2   Given $y = x^2e^{ax}$, find $\frac{d^n y}{dx^n}$ by Leibnitz's Formula.

Solution. Let $u = x^2$, and $v = e^{ax}$; then $\frac{du}{dx} = 2x$, $\frac{dv}{dx} = ae^{ax}$, $\frac{d^2 u}{dx^2} = 2x$, $\frac{d^2 v}{dx^2} = a^2 e^{ax}$, $\frac{d^3 u}{dx^3} = 0$, $\frac{d^3 v}{dx^3} = a^3 e^{ax}$, ..., $\frac{d^n u}{dx^n} = 0$, $\frac{d^n v}{dx^n} = a^n e^{ax}$. Substituting in (7.1), we get

$$\frac{d^n y}{dx^n} = x^2 a^n e^{ax} + 2na^{n - 1} x e^{ax} + n(n - 1)a^{n - 2} e^{ax} = a^{n - 2} e^{ax} [ x^2 a^2 + 2nax + n(n - 1) ].$$