The Product Rule 

1.0What is the Product Rule?

In simple terms, the Product Rule for differentiation is a special formula used to differentiate a function that is a product of two differentiable functions. For example, if you have a function h(x)=f(x)⋅g(x), where f(x) and g(x) are both differentiable functions, you cannot simply differentiate each part separately and multiply the results. That is a common mistake. Instead, you must apply the product rule to find the correct derivative, h′(x).

2.0The Product Rule Formula

If u=f(x) and v=g(x) are two differentiable functions, then the derivative of their product, uv, is given by:

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

In simpler notation, this can be written as:

(fg)' = f'g + fg'

This formula states that the derivative of the product of two functions is equal to (the first function multiplied by the derivative of the second) plus (the second function multiplied by the derivative of the first).

3.0Derivation of the Product Rule

While the formula itself is easy to apply, understanding its origin is key to a deeper conceptual grasp, which is vital for JEE. We can derive the product rule using the first principle of differentiation (the limit definition).

Let h(x)=f(x)g(x). The derivative of h(x) is given by:

$h^{\prime }(x)={\lim }_{t\to 0}\frac{h(x+t)-h(x)}{t}$

Substituting h(x)=f(x)g(x) into the equation:

$h^{\prime }(x)={\lim }_{t\to 0}\frac{f(x+t)g(x+t)-f(x)g(x)}{t}$

This is a bit tricky to solve directly. The key insight is to add and subtract a term in the numerator. We add and subtract f(x)g(x+t):

$h^{\prime }(x)={\lim }_{t\to 0}\frac{f(x+t)g(x+t)-f(x)g(x+t)+f(x)g(x+t)-f(x)g(x)}{t}$

Now, we can rearrange the terms and factor them:

$h^{\prime }(x)={\lim }_{t\to 0}\left[\frac{g(x+t)[f(x+t)-f(x)]}{t}+\frac{f(x)[g(x+t)-g(x)]}{t}\right]$

$h^{\prime }(x)={\lim }_{t\to 0}\left[g(x+t)\left(\frac{f(x+t)-f(x)}{t}\right)+f(x)\left(\frac{g(x+t)-g(x)}{t}\right)\right]$

Using the properties of limits, we can split this into two separate limits:

$h^{\prime }(x)={\lim }_{t\to 0}g(x+t)\cdot {\lim }_{t\to 0}\left(\frac{f(x+t)-f(x)}{t}\right)+{\lim }_{t\to 0}f(x)\cdot {\lim }_{t\to 0}\left(\frac{g(x+t)-g(x)}{t}\right)$

As t→0, g(x+t)→g(x), and the terms in the parentheses are the definitions of the derivatives of f(x) and g(x). Thus, we get:

h′(x)=g(x)f′(x)+f(x)g′(x)

This derivation solidifies the formula and is a great way to remember it.

4.0Step-by-Step Method to Apply the Product Rule

1. Identify the functions: Clearly identify the two functions, u and v, that are being multiplied. For example, if $y=x^2\sin (x),\text{let }u=x^2\text{ and }v=\sin (x)$
2. Find the derivatives: Calculate the derivative of each individual function. Find $\frac{du}{dx}\text{ and }\frac{dv}{dx}$
3. Apply the formula: Substitute the original functions and their derivatives into the product rule formula: $u\frac{dv}{dx}+v\frac{du}{dx}$.
4. Simplify: Perform the necessary algebraic simplification to get the final answer.

5.0Solved Examples on the Product Rule

Example 1: Basic Application with Polynomials

Problem: Find the derivative of $y=(x^2+1)(x^3-x).$

Solution:

1. Identify the functions: $\text{Let }\breve{=}{x}^2+1\text{ and }\check{=}{x}^3-x$

Find the derivatives:

$\frac{du}{dx}=\frac{d}{dx}(x^2+1)=2x$

$\frac{dv}{dx}=\frac{d}{dx}(x^3-x)=3x^2-1$