product rule

1. Introduction

In differential calculus, we frequently encounter functions that are expressed as the product of two other functions.

For example:

$y=x^2\sin x$

Or:

$y=(x^2+3x)(x^3-2)$

Differentiating such expressions requires a special rule known as the Product Rule of Differentiation.

The product rule states that the derivative of the product of two differentiable functions is obtained by differentiating the first function and multiplying it by the second, then adding the first function multiplied by the derivative of the second.

Mathematically, if

$y=f(x)g(x)$

then

$dy/dx=f'(x)g(x)+f(x)g'(x)$

This can also be written as:

$d[f(x)g(x)]/dx=f'(x)g(x)+f(x)g'(x)$

But why does this rule work?

Why can’t we simply differentiate both functions and multiply their derivatives?

To understand the product rule properly, let us derive it from first principles and explore its geometrical meaning.

2. Why Do We Need the Product Rule?

Suppose we have two functions:

$f(x)=x^2$ $g(x)=x^3$

Their product is:

$y=f(x)g(x)$

Therefore,

$y=x^2\times x^3$

Using the laws of exponents:

$y=x^5$

Differentiating directly:

$dy/dx=5x^4$

Now consider differentiating the two original functions separately.

$f'(x)=2x$ $g'(x)=3x^2$

Multiplying their derivatives gives:

$f'(x)g'(x)=6x^3$

Clearly, this is not equal to the actual derivative.

Therefore, the derivative of a product cannot generally be obtained by simply multiplying the derivatives of its individual functions.

We need a different approach.

That approach is the product rule.

3. Deriving the Product Rule from First Principles

The derivative of a function is defined as the limit of its difference quotient.

For a function y = F(x), its derivative is:

$F'(x)=\lim_{h\to 0}(F(x+h)-F(x))/h$

Now suppose:

$F(x)=f(x)g(x)$

Replacing x with x + h gives:

$F(x+h)=f(x+h)g(x+h)$

Substituting these expressions into the definition of the derivative:

$F'(x)=\lim_{h\to 0}(f(x+h)g(x+h)-f(x)g(x))/h$

The important step is to rearrange the numerator.

We can add and subtract the expression:

$f(x)g(x+h)$

This gives:

$F'(x)=\lim_{h\to 0}(f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x))/h$

Grouping the terms:

$F'(x)=\lim_{h\to 0}[g(x+h)(f(x+h)-f(x))/h+f(x)(g(x+h)-g(x))/h]$

Separating the expressions:

$F'(x)=\lim_{h\to 0}[g(x+h)((f(x+h)-f(x))/h)+f(x)((g(x+h)-g(x))/h)]$

Using the limit laws and the continuity of differentiable functions:

$F'(x)=g(x)f'(x)+f(x)g'(x)$

Rearranging:

$F'(x)=f'(x)g(x)+f(x)g'(x)$

Therefore, we obtain the product rule:

$d[f(x)g(x)]/dx=f'(x)g(x)+f(x)g'(x)$

The product rule is therefore a direct consequence of the definition of the derivative.

4. Geometrical Interpretation of the Product Rule

The product rule can also be understood through a simple geometrical idea involving the area of a rectangle.

Suppose the length and width of a rectangle are represented by two functions:

$L=f(x)$ $W=g(x)$

The area of the rectangle is:

$A=f(x)g(x)$

Now imagine that x increases by a small amount, h.

As a result, both the length and width may change.

The new dimensions become:

$L_{\mathrm{new}}=f(x)+\Delta f$ $W_{\mathrm{new}}=g(x)+\Delta g$

The new area is:

$A_{\mathrm{new}}=(f(x)+\Delta f)(g(x)+\Delta g)$

Expanding:

$A_{\mathrm{new}}=f(x)g(x)+f(x)\Delta g+g(x)\Delta f+\Delta f\Delta g$

The change in area is therefore:

$\Delta A=f(x)\Delta g+g(x)\Delta f+\Delta f\Delta g$

Dividing by h:

$\Delta A/h=f(x)(\Delta g/h)+g(x)(\Delta f/h)+\Delta f\Delta g/h$

As h approaches zero, the product of the two small changes becomes negligible after division by h.

Taking the limit:

$dA/dx=f(x)g'(x)+g(x)f'(x)$

Thus:

$dA/dx=f'(x)g(x)+f(x)g'(x)$

This is precisely the product rule.

Understanding the Geometry

The change in the rectangle’s area can be understood through three contributions:

The area added because the width changes.
The area added because the length changes.
The small corner area created when both dimensions change.

As the change in x approaches zero, the third contribution becomes negligible compared with the first two.

This explains why the derivative contains two terms rather than just one.

5. Worked Examples

Let us apply the product rule to some common functions.

Example 1: Product of Two Polynomial Functions

Differentiate:

$y=x^2(x^3+2)$

Let:

$f(x)=x^2$ $g(x)=x^3+2$

Differentiating:

$f'(x)=2x$ $g'(x)=3x^2$

Applying the product rule:

$dy/dx=f'(x)g(x)+f(x)g'(x)$

Substituting:

$dy/dx=2x(x^3+2)+x^2(3x^2)$

Expanding:

$dy/dx=2x^4+4x+3x^4$

Therefore:

$dy/dx=5x^4+4x$

Example 2: Product of a Polynomial and a Trigonometric Function

Differentiate:

$y=x^2\sin x$

Let:

$f(x)=x^2$ $g(x)=\sin x$

Their derivatives are:

$f'(x)=2x$ $g'(x)=\cos x$

Applying the product rule:

$dy/dx=2x\sin x+x^2\cos x$

Therefore:

$dy/dx=2x\sin x+x^2\cos x$

Example 3: Product of Two Trigonometric Functions

Differentiate:

$y=\sin x\cos x$

Let:

$f(x)=\sin x$ $g(x)=\cos x$

Their derivatives are:

$f'(x)=\cos x$ $g'(x)=-\sin x$

Applying the product rule:

$dy/dx=\cos x\cos x+\sin x(-\sin x)$

Simplifying:

$dy/dx=\cos^2 x-\sin^2 x$

Using the double-angle identity:

$\cos 2x=\cos^2 x-\sin^2 x$

Therefore:

$dy/dx=\cos 2x$

This example demonstrates an interesting connection between differentiation and trigonometric identities.

6. Common Mistake: Multiplying the Derivatives

One of the most common mistakes students make is assuming:

$d[f(x)g(x)]/dx=f'(x)g'(x)$

This is incorrect.

The correct expression is:

$d[f(x)g(x)]/dx=f'(x)g(x)+f(x)g'(x)$

Remember:

Differentiate the first function while keeping the second unchanged. Then differentiate the second function while keeping the first unchanged. Finally, add both expressions.

This simple principle makes the product rule easier to remember and apply.

7. Conclusion

The product rule is one of the fundamental rules of differential calculus.

Rather than being an arbitrary formula to memorise, it follows naturally from the definition of the derivative.

Its geometrical interpretation also provides an intuitive explanation: when two dimensions of a rectangle change, the resulting change in area comes from the contribution of each changing dimension.

The essential formula is:

$d[f(x)g(x)]/dx=f'(x)g(x)+f(x)g'(x)$

Understanding the reasoning behind this formula makes it easier to apply the product rule to polynomial, trigonometric, exponential and logarithmic functions.

The broader lesson is that differentiation describes how quantities change, and when two quantities are multiplied, the change in their product depends on how each quantity changes individually.

The Product Rule of Differentiation: Understanding Why the Derivative of a Product Works

1. Introduction

2. Why Do We Need the Product Rule?

3. Deriving the Product Rule from First Principles

4. Geometrical Interpretation of the Product Rule

Understanding the Geometry

5. Worked Examples

Example 1: Product of Two Polynomial Functions

Example 2: Product of a Polynomial and a Trigonometric Function

Example 3: Product of Two Trigonometric Functions

6. Common Mistake: Multiplying the Derivatives

7. Conclusion

Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

product rule

1. Introduction

2. Why Do We Need the Product Rule?

3. Deriving the Product Rule from First Principles

4. Geometrical Interpretation of the Product Rule

Understanding the Geometry

5. Worked Examples

Example 1: Product of Two Polynomial Functions

Example 2: Product of a Polynomial and a Trigonometric Function

Example 3: Product of Two Trigonometric Functions

6. Common Mistake: Multiplying the Derivatives

7. Conclusion

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series