Product Rule in Derivatives

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.

[u(x)v(x)]' = u(x)v'(x) + u'(x)v(x)

When the derivative of the 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 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 found 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)

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) = lim_x→0{h(x + △x) - h(x)}/△x

= lim_x→0{a(x + △x).b(x + △x) - a(x).b(x)}/△x

= lim_x→0{a(x + △x).b(x + △x) - a(x).b(x + △x) - a(x).b(x)}/△x

= lim_x→0{[a(x + △x) - a(x)].b(x + △x) - a(x).[b(x + △x) - b(x)]}/△x

= lim_x→0{[a(x + △x) - a(x)].b(x + △x)}/△x - lim_x→0[a(x).[b(x + △x) - b(x)]}/△x

= {lim_x→0{a(x + △x) - a(x)}/△x}.{lim_x→0b(x + △x)} + {lim_x→0{b(x + △x) - b(x)}/△x}.{lim_x→0a(x + △x)}

= b(x).{lim_x→0{a(x + △x) - a(x)}/△x} + {lim_x→0{b(x + △x) - b(x)}/△x}.a(x)

Now,

• {lim_x→0{a(x + △x) - a(x)}/△x} = a'(x)
• {lim_x→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 vs Quotient Rule

Key difference between product and quotient rule is given as follows:

Solved Examples on Product Rule

Example 1: Find the derivative of the function y = e^x sinx

Solution:

y = e^x.sinx

By Using Product Rule

y′(x) = (e^xsinx)′

⇒ y′(x) = (e^x)′sinx + e^x(sinx)′ 

⇒ y′(x) = e^xsinx + e^x(cosx)

⇒ y′(x) = e^x(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 = x² 3^x.

Solution:

Given y = x²3^x

f(x) = x² and f'(x) = 2x

g(x) = 3^x and g'(x) =3^xlog3

Now

⇒ y' = f(x)g'(x) + f'(x)g(x)

⇒ y' = x²3^xlog3 + 2x3^x

⇒ y' = 3^xx(xlog3 + 2)

Example 4 : Find the differentiation of y = e^x(cosx - sinx).

y = e^x(cosx - sinx)

Let f(x) =e^x then f'(x) = e^x

and g(x) = cosx - sinx , then g' (x) = -sinx -cosx

So

y' = f(x) g'(x) + f'(x) g(x)

⇒ y' = e^x(-sinx - cosx ) + e^x (cosx - sinx)

⇒ y' = -2sinxe^x

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.

Related Articles:

• Derivatives
• Calculus in Maths
• Differential Calculus
• Differentiation and Integration Formula
• Advanced Differentiation
• Implicit Differentiation