Product Rule is the rule that is used to find the derivative of the 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 the functions.
Product Rule of the calculus is proved using the concept of limit and derivatives. In this article, we will learn about the Product rule, the product rule formula, its proof, examples, and others in detail in this article.
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)
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 + â–³x) – h(x)}/â–³x
= limx→0{a(x + â–³x).b(x + â–³x) – a(x).b(x)}/â–³x
= limx→0{a(x + â–³x).b(x + â–³x) – a(x).b(x + â–³x) – a(x).b(x)}/â–³x
= limx→0{[a(x + â–³x) – a(x)].b(x + â–³x) – a(x).[b(x + â–³x) – b(x)]}/â–³x
= limx→0{[a(x + â–³x) – a(x)].b(x + â–³x)}/â–³x – limx→0[a(x).[b(x + â–³x) – b(x)]}/â–³x
= {limx→0{a(x + â–³x) – a(x)}/â–³x}.{limx→0b(x + â–³x)} + {limx→0{b(x + â–³x) – b(x)}/â–³x}.{limx→0a(x + â–³x)}
= b(x).{limx→0{a(x + â–³x) – a(x)}/â–³x} + {limx→0{b(x + â–³x) – b(x)}/â–³x}.a(x)
Now,
- {limx→0{a(x + â–³x) – a(x)}/â–³x} = a'(x)
- {limx→0{b(x + â–³x) – b(x)}/â–³x} = 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)
Applying 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.
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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)′
= (ex)′sinx + ex(sinx)′
= exsinx + ex(cosx)
= ex(sinx + cosx)
Example 2: Find the derivative of the function Z= (y³ + 2y²-y)(eʸ – 1 ) .
Z=(y³ + 2y²-y)(eʸ – 1 )
Zʹ(x)=((y³ + 2y²-y)(eʸ – 1 ) ) ʹ
=(y³ + 2y²-y)(eʸ – 1 ) ʹ + (y³ + 2y²-y) ʹ (eʸ – 1 )
=(y³ + 2y²-y)(eʸ ) + ( 3y² + 4y -1 )(eʸ – 1 )
=(y³ + 5y²-3y-1 )(eʸ) -( 3y² + 4y -1 )
Example 3 : Find the derivative of the function y=x23x .
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)
=x23xlog3 + 2×3x
=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)
=ex(-sinx – cosx ) + ex (cosx – sinx)
=-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)
=u(vw’ + v’w ) + u'(vw)
Example 6 : Find the derivative of y =sinx.cosx
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)
=sinx . (-sinx) + cosx . (cosx)
=cos2x – sin2x
Using Identity cos2x = cos2x – sin2 x
y’ = cos2x
FAQs on Product Rule
What is Product Rule of Differentiation in Calculus?
The product rule of the differentiation is the rule used in calculus to find the differentiation of the product of the two functions.
What is Product Rule Formula?
The product rule formula is the formula that is used to find the differentiation of two function, suppose we have to find the differentiation of f(x) = h(x).g(x) such that,
f'(x) = h(x).g'(x) + g(x).h'(x)
What is the Use of Product Rule in Differentiation?
The product rule in differentiation is used for various purposes,
- It is used to find the differentiation of the function that are expressed as the product of two functions.
- It is used to find the rate, maxima, minima, etc and others in detail, etc.
What is Quotient Rule?
The quotient rule in the differentiation is used to find the differentiation of the function that are expressed as the division of two functions,
Suppose we have a function f(x) = g(x)/h(x) then the differentiation of f(x) is found as,
f'(x) = {h(x).g'(x) – g(x).h'(x)}/ {h(x)2}
How the product rule is related to chain Rule ?
The chain rule is used for differentiating composite functions, where one function is nested inside another. The product rule can be seen as a special case of the chain rule where the composite function is simply multiplication.
Are there any limitation of product rule ?
The product rule assumes that both functions being multiplied are differentiable. If one of the functions isn’t differentiable at a certain point, the product rule can’t be applied directly at that point.
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