Leibnitz Theorem on Successive Differentiation: Solved Problems

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.

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_nv + ⁿc₁ u_n-1v₁ + ⁿc₂ u_n-2v₂ + … + uv_n.

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

Solved Examples

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

Solution:

Given that y = x^n-1 logx.

Differentiating both sides with respect to x, we obtain that

y₁ = (n-1) x^n-2 logx + x^n-1 ⋅ $\dfrac{1}{x}$

⇒ xy₁ = (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:

y1x = (n-1)y + xn-1.

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

(y₁x)_n-1 = (n-1)y_n-1 + (x^n-1)_n-1

⇒ y_nx + ^n-1c₁ y_n-1 ⋅1 = (n-1)y_n-1 + (n-1)! as the n-th derivative of xⁿ is equal to n!

⇒ y_nx = (n-1)!

⇒ y_n = $\dfrac{(n-1)!}{x}$ (Proved.)

Q2: If y = tan-1x, then show that (1+x2)yn+1 +2nxyn +n(n-1)yn-1 = 0.

Solution:

We have y = tan^-1x.

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

y₁ = $\dfrac{1}{1+x^2}$

⇒ (1+x²) y₁ = 1

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

[y₁(1+x²)]_n = (1)_n

⇒ y_n+1(1+x²) + ⁿc₁ y_n ⋅2x + ⁿc₂ y_n-1 ⋅2 =0

⇒ (1+x²)y_n+1 + 2nxy_n + n(n-1)y_n-1 = 0 (Proved.)

Also Read: Beta and Gamma Functions

Q3: If y = em sin-1x, then show that (1-x2)yn+2 – (2n+1)xyn+1 -(n2+m2)yn = 0.

Solution:

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

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

y₁ = e^{m sin-1x} × $\dfrac{m}{\sqrt{1-x^2}}$ = $\dfrac{my}{\sqrt{1-x^2}}$

⇒ y₁²(1-x²) = m²y².

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

2y₁ y₂(1-x²) +y₁² ⋅(-2x) = m²⋅2yy₁

⇒ y₂ (1-x²) – xy₁ = m²y (cancelling 2y₁)

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

y_n+2 (1-x²) + ⁿc₁ y_n+1 ⋅(-2x) + ⁿc₁ y_n ⋅(-2) – {y_n+1x + ⁿc₁ y_n ⋅1} = m²y_n

⇒ (1-x²)y_n+2 – 2nxy_n+1 – n(n-1)y_n – xy_n+1 – ny_n = m²y_n

⇒ (1-x²)y_n+2 – (2n+1)xy_n+1 -(n²+m²)y_n = 0 (Proved.)