Product Rule in Derivatives
The Product Rule is the rule that is used to find the derivative of a function that is expressed as the product of two functions. The product rule in calculus is the fundamental rule and is used to find the derivative of functions.
The Product Rule of calculus is proved using the concept of limits and derivatives. In this article, we will learn about the Product Rule, the product rule formula, its proof, examples, and others in detail.

What is Product Rule?
When the derivative of product of two or more functions is to be taken, the product rule is applied. The product rule states that if a function is the product of the two functions then the derivative of the function is the sum of the product of the first function and the derivative of the second function, with the product of the second function and the derivative of the first function.
For any given function that is the product of the two functions,
d/dx{f(x)·g(x)} = [g(x) × f'(x) + f(x) × g'(x)]
Product Rule Formula
The product rule formula in calculus is the formula that gives the way to find the differentiation of two functions and the formula for the product rule formula is given as,
Suppose we have f(x) = u(x).v(x) then the differentiation of f(x) is find as,
d/dx{u(x)·v(x)} = [v(x) × u'(x) + u(x) × v'(x)]
Where,
- u(x) and v(x) are the differential functions
- u'(x) is the derivative of u(x)
- v'(x) is the derivative of v(x)
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How to Apply Product Rule in Differentiation
Product rule is applied to the product of the function, follow the steps discuss below,
Step 1: Identify the function f(x) and g(x)
Step 2: Find the derivative functions f'(x) and g'(x)
Step 3: Use the formula,
d/dx{f(x).g(x)} = f(x).g'(x) + f'(x).g(x)
Then use the formula to get the required differentiation.
Product Rule Example
Let's consider an example for better understanding,
Example : Find the derivative of y = sinx . cosx.
Solution:
Given y =sinx.cosx
Let f(x) =sinx and g(x) =cosx
then
f'(x) =cosx and g'(x) = -sinx
So y' = f(x)g'(x) + f'(x)g(x)
⇒ y' = sinx . (-sinx) + cosx . (cosx)
⇒ y' = cos2x - sin2x
Using Identity cos2x = cos2x - sin2 x
y' = cos2x
Derivation of Product Rule Formula
Let us take two functions a(x) and b(x). So, the Product rule arrives when you multiply the first function a(x) with the derivative of the second function b(x) plus the derivative of the first function a(x) multiplied by the second function b(x). Thus we make the product rule as,
(ab)' = a'b + ab'
In Leibniz's notation, it is written as,
d/dx(u.v) = du/dx.(v) + (u).dv/dx
This formula can be proved by two methods,
- Using First Principle
- Using Chain Rule
Now let's prove the same by both methods,
Product Rule Formula Using First Principle Proof
Using the first principle of the derivative we can easily prove the product rule as, suppose we have a function h(x) = a(x).b(x) then its differentiation is found using,
h'(x) = limx→0{h(x + △x) - h(x)}/△x
= limx→0{a(x + △x).b(x + △x) - a(x).b(x)}/△x
= limx→0{a(x + △x).b(x + △x) - a(x).b(x + △x) - a(x).b(x)}/△x
= limx→0{[a(x + △x) - a(x)].b(x + △x) - a(x).[b(x + △x) - b(x)]}/△x
= limx→0{[a(x + △x) - a(x)].b(x + △x)}/△x - limx→0[a(x).[b(x + △x) - b(x)]}/△x
= {limx→0{a(x + △x) - a(x)}/△x}.{limx→0b(x + △x)} + {limx→0{b(x + △x) - b(x)}/△x}.{limx→0a(x + △x)}
= b(x).{limx→0{a(x + △x) - a(x)}/△x} + {limx→0{b(x + △x) - b(x)}/△x}.a(x)
Now,
- {limx→0{a(x + △x) - a(x)}/△x} = a'(x)
- {limx→0{b(x + △x) - b(x)}/△x} = b'(x)
h'(x) = b(x).a'(x) + a(x).b'(x)
Thus, the product rule is proved
Product Rule Formula Using Chain Rule Proof
Using the Chain Rule of the derivative we can easily prove the product rule as, suppose we have a function h(x) = a(x).b(x) then its differentiation is found using,
d/dx.{h(x)} = d/dx.{a(x).b(x)} = d/dx.{a.b}
= {d(a.b)/da}.{da/dx} + {d(a.b)/db}.{db/dx}
= b.{da/dx} + a{db/dx}
= a'.b + a.b'
Thus, the product rule is proved.
Product Rule for Products of More Than Two Functions
Product rule for more than two functions is simply found using the product of two functions. And then applying the product rule again,
d(a.b.c)/dx = da/dx.(b.c) + a.(db/dx).c + a.b.(dc/dx)(d{a.b.c}/dx)
= da./dx(b.c) + a.(db/dx).c + a.b.(dc/dx)
Product and Quotient Rule
Key difference between product and quotient rule is given as follows:
Aspect | Product Rule | Quotient Rule |
---|---|---|
Definition | Used to find the derivative of the product of two functions. | Used to find the derivative of the quotient of two functions. |
Formula | If u(x) and v(x) are functions, the product rule is (uv)′ = u′v + uv′. | If u(x) and v(x) are functions, the quotient rule is (u/v)' = (u'v -uv')/v2 |
Application | Applicable when differentiating the product of two functions. | Applicable when differentiating the quotient (or division) of two functions. |
Complexity | Generally simpler and involves addition. | Generally more complex and involves subtraction and division. |
Components Needed | Derivatives of the individual functions u(x) and v(x). | Derivatives of the individual functions u(x) and v(x) as well as the function v(x) itself. |
Common Use Cases | Physics (e.g., product of displacement and velocity), economics (e.g., product of cost and quantity). | Physics (e.g., ratio of forces), economics (e.g., average cost function). |
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Solved Examples on Product Rule
Example 1: Find the derivative of the function y = exsinx
Solution:
y = ex.sinx
By Using Product Rule
y′(x) = (exsinx)′
⇒ y′(x) = (ex)′sinx + ex(sinx)′
⇒ y′(x) = exsinx + ex(cosx)
⇒ y′(x) = ex(sinx + cosx)
Example 2: Find the derivative of the function Z= (y³ + 2y²-y)(eʸ - 1 ).
Solution:
Z = (y³ + 2y²-y)(eʸ - 1 )
Thus, Zʹ(x) = ((y³ + 2y²-y)(eʸ - 1 ) ) ʹ
⇒ Zʹ(x) =(y³ + 2y²-y)(eʸ - 1 ) ʹ + (y³ + 2y²-y) ʹ (eʸ - 1 )
⇒ Zʹ(x) =(y³ + 2y²-y)(eʸ ) + ( 3y² + 4y -1 )(eʸ - 1 )
⇒ Zʹ(x) =(y³ + 5y²-3y-1 )(eʸ) -( 3y² + 4y -1 )
Example 3 : Find the derivative of the function y=x23x.
Solution:
Given y = x23x
f(x) = x2 and f'(x) = 2x
g(x) = 3x and g'(x) =3xlog3
Now
⇒ y' = f(x)g'(x) + f'(x)g(x)
⇒ y' = x23xlog3 + 2x3x
⇒ y' = 3xx(xlog3 + 2)
Example 4 : Find the differentiation of y=ex(cosx-sinx).
y = ex(cosx-sinx)
Let f(x) =ex then f'(x) = ex
and g(x) =cosx - sinx , then g' (x) =-sinx -cosx
So
y' = f(x) g'(x) + f'(x) g(x)
⇒ y' = ex(-sinx - cosx ) + ex (cosx - sinx)
⇒ y' = -2sinxex
Example 5 : Find the derivative of y =u.v.w , where u,v,w,y are function of x.
Given y = u . v . w
Let f(x) = u and g(x) = v.w
then f'(x) = u' and g'(x) =vw' + v'w
So, y' = f(x)g'(x) + f'(x) g(x)
⇒ y' =u(vw' + v'w ) + u'(vw)
Product Rule in Derivatives Practice Problems
1. Find the derivative of f(x) = x²sin(x) using the product rule.
2. Differentiate g(x) = (x³ + 2x)(4x - 1) using the product rule.
3. Find d/dx [e^x · ln(x)] using the product rule.
4. Determine the derivative of h(x) = x(x² + 1)³ using the product rule.
5. Calculate the derivative of f(x) = (x + 2)(x² - 3x + 1) using the product rule.
Summary
The Product Rule is a fundamental principle in calculus used to find the derivative of a product of two or more functions. It states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. Mathematically, if f(x) and g(x) are two differentiable functions, then the derivative of their product is given by: d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x). This rule is essential for differentiating complex functions that involve multiplication and is widely applied in various fields of mathematics, physics, and engineering where rates of change of composite quantities need to be calculated.