The Leibnitz’s theorem is used to find the n-th order derivative of a product function. This theorem is a generalisation of the product rule of differentiation. In this article, we will learn about Leibnitz’s theorem and solve few problems to learn how to use this theorem.

Table of Contents

## Statement of Leibnitz’s Theorem

Let u and v be two continuously differentiable functions.Then the n-th order derivative of uv is given by the formula below:

(uv)_{n} = u_{n}v + ^{n}c_{1} u_{n-1}v_{1} + ^{n}c_{2} u_{n-2}v_{2} + … + uv_{n}.

In the next section, we will find n-th order derivatives of product functions.

## Solved Examples

Q1: If y = x^{n-1} logx, then show that y_{n} = $\dfrac{(n-1)!}{x}$ |

**Solution:**

Given that y = x^{n-1} logx.

Differentiating both sides with respect to x, we obtain that

y_{1} = (n-1) x^{n-2 }logx + x^{n-1} ⋅ $\dfrac{1}{x}$

⇒ xy_{1} = (n-1) x^{n-1 }logx + x^{n-1} (multiplying by x on both sides)

As y = x^{n-1} logx, we deduce the following:

y_{1}x = (n-1)y + x^{n-1}. |

Now, differentiating (n-1)-times w.r.t x using Leibnitz’s theorem it follows that

(y_{1}x)_{n-1} = (n-1)y_{n-1} + (x^{n-1})_{n-1}

⇒ y_{n}x + ^{n-1}c_{1} y_{n-1} ⋅1 = (n-1)y_{n-1} + (n-1)! as the n-th derivative of x^{n} is equal to n!

⇒ y_{n}x = (n-1)!

⇒ y_{n} = $\dfrac{(n-1)!}{x}$ **(Proved.)**

Q2: If y = tan^{-1}x, then show that (1+x^{2})y_{n+1} +2nxy_{n} +n(n-1)y_{n-1} = 0. |

**Solution:**

We have y = tan^{-1}x.

Differentiating both sides w.r.t x, it follows that

y_{1} = $\dfrac{1}{1+x^2}$

⇒ (1+x^{2}) y_{1} = 1

Differentiating n-times using Leibnitz’s theorem, we obtain that

[y_{1}(1+x^{2})]_{n} = (1)_{n}

⇒ y_{n+1}(1+x^{2}) + ^{n}c_{1} y_{n} ⋅2x + ^{n}c_{2} y_{n-1} ⋅2 =0

⇒ (1+x^{2})y_{n+1} + 2nxy_{n} + n(n-1)y_{n-1} = 0 **(Proved.)**

**Also Read:** Beta and Gamma Functions

Q3: If y = e^{m sin-1x}, then show that (1-x^{2})y_{n+2} – (2n+1)xy_{n+1} -(n^{2}+m^{2})y_{n} = 0. |

**Solution:**

Given y = e^{m sin-1x}.

Differentiating both sides w.r.t x, we have that

y_{1} = e^{m sin-1x} × $\dfrac{m}{\sqrt{1-x^2}}$ = $\dfrac{my}{\sqrt{1-x^2}}$

⇒ y_{1}^{2}(1-x^{2}) = m^{2}y^{2}.

Again differentiate w.r.t x. So we obtain that

2y_{1} y_{2}(1-x^{2}) +y_{1}^{2} ⋅(-2x) = m^{2}⋅2yy_{1}

⇒ y_{2} (1-x^{2}) – xy_{1} = m^{2}y (cancelling 2y_{1})

Now differentiating n-times w.r.t x using Leibnitz’s theorem, it follows that

y_{n+2} (1-x^{2}) + ^{n}c_{1} y_{n+1} ⋅(-2x) + ^{n}c_{1} y_{n} ⋅(-2) – {y_{n+1}x + ^{n}c_{1} y_{n} ⋅1} = m^{2}y_{n}

⇒ (1-x^{2})y_{n+2} – 2nxy_{n+1} – n(n-1)y_{n} – xy_{n+1} – ny_{n} = m^{2}y_{n}

⇒ (1-x^{2})y_{n+2} – (2n+1)xy_{n+1} -(n^{2}+m^{2})y_{n} = 0 **(Proved.)**

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.