Derivatives of different orders — Leibniz’ Rule: IITJEE Maths training

Let a function $y=f(x)$ be differentiable on some interval $[a,b]$ . Generally speaking, the values of the derivative $f^{'}(x)$ depend on x, which is to say that the derivative $f^{'}(x)$ is also a function of x. Differentiating this function, we obtain the so-called second derivative of the function $f(x)$ .

The derivative of a first derivative is called a derivative of the second order or the second derivative of the original function and is denoted by the symbol $y^{''}$ or $f^{''}(x)$ : $y^{''}=(y^{'})^{'} = f^{''}(x)$

For example, if $y=x^{5}$ , then $y^{'}=5x^{4}$ and $y^{''} = (5x^{4})^{'}=20x^{3}$

The derivative of the second derivative is called a derivative of the third order or the third derivative and is denoted by $y^{'''}$ or $f^{'''}(x)$ .

Generally, a derivative of the nth order of a function $f(x)$ is called the derivative (first order) of the derivative of the $(n-1)$ th order and is denoted by the symbol $y^{(n)}$ or $f^{(n)}(x)$ :

$y^{(n)} = (y^{(n-1)})^{'}=f^{(n)}(x)$

(Note: the order of the derivative is taken in parentheses so as to avoid confusion with the exponent of a power.)

Derivatives of the fourth, fifth and higher orders are also denoted by Roman numerals: $y^{IV}$ , $y^{V}$ , $y^{VI}$ , $\ldots$ . Here, the order of the derivative may be written without brackets. For instance, if $y=x^{5}$ , $y^{'}=5x^{4}$ , $y^{''}=20x^{3}$ , $y^{'''}=60x^{2}$ , $y^{IV}=y^{(4)}=120x$ , $y^{V}=y^{(5)}=100$ , $y^{(6)}=y^{(7)}=\ldots = 0$

Example 1:

Given a function $y=e^{kx}$ , where k is a constant, find the expression of its derivative of any order n.

Solution 1:

$y^{'}=ke^{kx}$ , $y^{''}=k^{2}e^{kx}$ , $y^{(n)}=k^{n}e^{kx}$

Example 2:

$y=\sin{x}$ . Find $y^{(n)}$ .

Solution 2:

$y^{'}=\cos{x}=\sin(x+\frac{\pi}{2})$

$y^{''}=-\sin{x}=\sin(x+2\frac{\pi}{2})$

$y^{'''}=-\cos{x}=\sin(x+3\frac{\pi}{2})$

$y^{IV}=\sin{x}=\sin(x+4\frac{\pi}{2})$

$\vdots$

$y^{(n)}=\sin(x+n\frac{\pi}{2})$

In similar fashion, we can also derive the formulae for the derivatives of any order of certain other elementary functions. You can find yourself the formulae for derivatives of the $n^{th}$ order of the functions $y=x^{k}$ , $y=\cos{x}$ , $y=\ln (x)$ .

Let us derive a formula called the Leibniz rule that will enable us to calculate the $n^{th}$ derivative of the product of two functions $u(x), v(x)$ . To obtain this formula, let us first find several derivatives and then establish the general rule for finding the derivative of any order:

$y=uv$

$y^{'}=u^{'}v+uv^{'}$

$y^{''}=u^{''}v+u^{'}v^{'}+u^{'}v^{'}+uv^{''}=u^{''}v+2u^{'}v^{'}+uv^{''}$

$y^{'''}=u^{'''}v+u^{''}v^{'}+2u^{''}v^{'}+2u^{'}v^{''}+u^{'}v^{''}+uv^{'''}$

which in turn equals $u^{'''}v+3u^{''}v^{'}+3u^{'}v^{''}+uv^{'''}$

$y^{IV}=u^{IV}v+4u^{'''}v+6u^{''}v^{''}+4u^{'}v^{'''}+uv^{IV}$

The rule for forming derivatives holds for the derivative of any order and obviously consists in the following:

The expression $(u+v)^{n}$ is expanded by the binomial theorem, and in the expansion obtained the exponents of the powers of u and v are replaced by indices that are the orders of the derivatives, and the zero powers $(u^{0}=v^{0}=1)$ in the end terms of the expansion are replaced by the function themselves (that is, “derivatives of zero order”):

$y^{(n)}=(uv)^{(n)}=u^{(n)}v+nu^{(n-1)}v^{'}+\frac{n(n-1)}{1.2}u^{(n-2)}v^{''}+\ldots + uv^{(n)}$

This is the Leibniz Rule.

A rigorous proof of this formula may be carried out by the method of complete mathematical induction (in other words, to prove that if this formula holds for the nth order, it will also hold for the order $n+1$ ).