Leibniz’s Theorem 

Leibniz's theorem is an important concept in maths that has special application in calculus and vector algebra which can teach one how to calculate the derivative of a product of two functions. This theorem finds application in physics, engineering, and computer science basically because it is an easy way of doing complicated calculations. 

1.0What is Leibniz’s Theorem?

Leibniz's theorem deals with the differentiation of a product of two functions. If we have two functions, Leibniz's theorem provides a rule to find their derivatives when multiplied.

It states that: 

Definition 1: It states that if

$f(x)\text{ and }g(x)$ are differentiable functions, then the derivative of their product is given by:

$\frac{d}{dx}[f(x)\cdot g(x)]=f^{\prime }(x)\cdot g(x)+f(x)\cdot g^{\prime }(x)$

Definition 2: If f(x) and g(x) two functions are n-times differentiable, then the nth derivative of their product f(x), g(x) is given by:

$(fg{)}^n={\sum }_{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{n-k}{g}^k$

Here,

$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$

Leibniz's theorem formula is known as the product rule and is a special case of Leibniz's theorem, a more general statement about functions raised to powers and multiplied together. 

2.0Leibniz Theorem Proof

The proof of the Leibniz theorem is typically done by induction. The formula is true for the n-th derivative and then deduces that it is true for the (n + 1)-th derivative by using the product rule repeatedly. The generalised form of Leibniz theorem is: 

$\frac{d}{dx}[f_1(x).f_2(x)\dots f_n(x)={\sum }_{k=1}^n(f_1(x)\dots \dots ...f_{\mathrm{k}-1(x)}$

Step 1: Proof for the product of Two Functions 

Let’s consider the case for two functions f₁(x) and f₂(x). The product rule says that: 

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\right]=f_1^{\prime }(x)\cdot f_2(x)+f_1(x)\cdot f_2^{\prime }(x)$

Step 2: Proof for the product of three functions 

differentiate the product f₁(x)⋅f₂(x) as a whole, and treat f₃(x) as a constant:

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\cdot f_3(x)\right]=\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\right]\cdot f_3(x)$

Use the product rule on f₁(x).f₂(x): 

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\right]={f^{\prime }}_1(x)\cdot f_2(x)+f_1(x)\cdot f_2^{\prime }(x)$

Hence, the derivative becomes: 

f₁’(x).f₂(x).f₃(x) + f₁.f₂’(x).f₃(x)

Now, the final equation will be: 

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\cdot f_3(x)\right]=f_1^{\prime }(x)\cdot f_2(x)\cdot f_3(x)+f_1(x)\cdot f_2^{\prime }(x)\cdot f_3(x)+f_1(x)\cdot f_2(x)\cdot f_3^{\prime }(x)$

Step 3: General Proof using Mathematical induction

The Leibniz theorem can be proved with the help of Mathematical induction for n-terms. Here’s how it works: 

Base Case: 

For n=2, 

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\right]=f_1^{\prime }(x)\cdot f_2(x)+f_1(x)\cdot f_2^{\prime }(x)$

So, the base case holds.

Inductive Step:

Assume that the Leibniz theorem holds for the product of 

n function, i.e., for functions,

$f(x),f_2(x),\dots ,f_n(x)$, the derivative is:

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\dots ..f_n(x)\right]={\sum }_{k=1}^n\left(f_1(x)\dots .f_{k-1}(x)\cdot f_1(x)\cdot f_k+f_1(x)\dots f_n(x)\right)$

Now, let’s prove it for n+1 functions. Consider the product of n+1 functions

$f_1(x),f_2(x),\dots f_n(x),f_n+f_1(x)$. We want to differentiate:

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\dots ...f_n(x)\cdot f_n+1(x)\right]$

We can treat the product of the first n functions

$f_1(x)\cdot f_2(x)\dots ..f_n(x)$as a single function, say

$g(x)+f_1(x)\cdot f_2(x)\dots ...f_n(x)$. Then, the derivative of the product becomes:

$\frac{d}{dx}\left[g(x)\cdot f_n+1(x)\right]=g^{\prime }(x)\cdot f_n+1(x)+g(x)\cdot f_n^{\prime }+1(x)$

By applying the inductive hypothesis g(x), we know that:

$g^{\prime }(x)={\sum }_{k=1}^n\left(f_1(x)\dots f_{k-1}(x)\cdot f_k^{\prime }(x)\cdot f_{k+1}(x)\dots \dots f_n(x)\right)$

Thus, the derivative of the product of n+1 functions is:

$\frac{d}{dx}\left[f_1(x)\cdot f_2(x)\dots f_n(x)\cdot f_{n+1}(x)\right]={\sum }_{k=1}^n\left(f_1(x)\dots ..f_{k-1(x)}\cdot f_k^{\prime }(x)\cdot f_{k+1}(x)\dots ..f_n(x)\right)+\left(f_1(x)\dots f_n(x)\cdot f_{n+1}^{\prime }(x)\right)$

Thus, the inductive step is true, and by induction, the Leibniz theorem holds for any number of functions.

3.0Leibniz Theorem Examples

To Get a better understanding of the theorem, let’s see some Leibniz Theorem Solved Examples: 

Example 1: Differentiate

$f(x)=\left(x^2+3x\right)\left(5x^2-2x\right)$using Leibniz’s rule. 

Solution: 

$f(x)=\left(x^2+3x\right)\left(5x^2-2x\right)$

In the given function use each factor as a separate function. 

$u(x)=x^2+3x\text{ and }v(x)=5x^2-2x$

By applying the product rule (Leibniz Rule): 

$\frac{d}{dx}[u(x)\cdot v(x)]=u^{\prime }(x)\cdot v(x)+u(x)\cdot v^{\prime }(x)$

Differentiate u(x) and v(x) 

$u^{\prime }(x)=2x+3\text{ and }{v}^{\prime }(x)=10x-2$

Now, apply the product rule,

$f^{\prime }(x)=\frac{d}{dx}\left[\left(x^2+3x\right)\left(5x^2-2x\right)\right]=(2x+3)\left(5x^2-2x\right)+\left(x^2+3x\right)(10x-2)$

$f^{\prime }(x)=2x\left(5x^2-2x\right)+3\left(5x^2-2x\right)+x^2(10x-2)+3x(10x-2)$

$f^{\prime }(x)=10x^3+11x-6x+10x^3+28x^2-6x$

$f^{\prime }(x)=20x^3+39x^2-12x$

Example 2: Find the derivative of

$f(x)=x^2\cdot e^x$ with the help of Leibniz’s Theorem. 

Solution: we have,

$f(x)=x^2\cdot e^x$

$u^{\prime }(x)=2x\text{ and }{v}^{\prime }(x)=e^x$

Applying the theorem,

$f^{\prime }(x)=\frac{d}{dx}[u(x)\cdot v(x)]=u^{\prime }(x)\cdot v(x)+u(x)\cdot v^{\prime }(x)$

$f^{\prime }(x)=(2x)\left(e^x\right)+\left(x^2\right)\left(e^x\right)$

$f^{\prime }(x)=2x{e}^x+x^2{e}^x$

$f^{\prime }(x)=e^x\left(2x+x^2\right)$

Example 3: Find the 4th derivative of f(x) = (x² + 3x + 5)e^2x

Solution: Using the Leibniz's theorem, the nth derivative of the product f(x) = u(x)v(x) is given by: 

$(u(x).v(x){)}^n={\sum }_{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})u(x{)}^{k}v(x{)}^{n-k}$

Let u(x) = x² + 3x + 5 and v(x) = e^2x

Computing the required derivatives of u(x) and v(x)

• u(x) = x² + 3x + 5

u'(x) = 2x+3, u"(x) = 2, u'" = 0, and u"" = 0

• v(x) = e^2x

v'(x) = 2e^2x, v"(x) = 4e^2x, v'"(x) = 8e^2x, and v""(x) = 16e^2x

Applying n = 4 in Leibniz’s theorem

$(u(x).v(x){)}^4={\sum }_{k=0}^4(\genfrac{}{}{0ex}{}{4}{k})u(x{)}^{k}v(x{)}^{4-k}$

Now, compute each term:

For k = 0 

$\frac{4!}{0!(4-0)!}\times u^{{}^{\prime \prime \prime \prime }}(x)\times v(x)=10e^{2x}=0$

For k = 1 

$\frac{4!}{1!(4-1)!}\times u^{{}^{\prime \prime \prime }}(x)\times v^{\prime }(x)=402e^{2x}=0$

For k = 2 

$\frac{4!}{2!(4-2)!}\times u{"}^{}(x)\times v"(x)=624e^{2x}=48e^{2x}$

For k = 3 

$\frac{4!}{3!(4-3)!}\times u^{\prime }(x)\times v^{\prime \prime \prime }(x)=4(2x+3)8e^{2x}=32(2x+3)e^{2x}$

For k = 4: 

$\frac{4!}{4!(4-4)!}\times u(x)\times v^{\prime \prime \prime \prime }(x)=1\times (x^2+3x+5)\times 16e^{2x}=16(x^2+3x+5)e^{2x}$

Adding up all the values: 

$(u(x).v(x){)}^4=0+0+48e^{2x}+32(2x+3)e^{2x}+16(x^2+3x+5)e^{2}x$

$(u(x).v(x){)}^4=48e^{2x}+(64x+96)e^{2x}+(16x^2+48x+80)e^{2x}$

$(u(x).v(x){)}^4=16x^2{e}^{2x}+(64x+48x)e^{2x}+(48+96+80)e^{2x}$

$(u(x).v(x){)}^4=16x^2{e}^{2x}+112x{e}^{2x}+224e^{2x}$

So, the 4th derivative is:

$[f(x){]}^4=16x^2{e}^{2x}+112x{e}^{2x}+224e^{2x}$