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Product Rule of Differentiation with Examples

The steps to prove the product rule of differentiation are presented along with examples, exercises and solutions.

Definition of the Derivative of a Function

The derivative f(x) of the function f(x) is defined as
f(x)=limh0f(x+h)f(x)h(1)


Derivative of the Product of two Functions

Let function f(x) be given by the product of two functions u(x) and v(x) written as
f(x)=u(x)v(x)
Using the definition of the derivative in (1), the derivative of f(x)=u(x)v(x) is given by

f(x)=limh0u(x+h)v(x+h)u(x)v(x)h(2)

Subtracting and adding the same quantity to the denominator of (2) above does not change it.
Subtract and add u(x)v(x+h) in the numerator of (2) above and write

f(x)=limh0u(x+h)v(x+h)u(x)v(x+h)u(x)v(x)+u(x)v(x+h)h(3)

Split (3) and rewrite it as follows
f(x)=limh0(u(x+h)v(x+h)u(x)v(x+h)h+u(x)v(x+h)u(x)v(x)h)(4)

From the properties of limits , the limit of a sum is equal to the sum of the limits, hence
f(x)=limh0u(x+h)v(x+h)u(x)v(x+h)h+limh0(u(x)v(x+h)u(x)v(x)h)(5)

Use factoring to rewrite the above as
f(x)=limh0v(x+h)u(x+h)u(x)h+limh0u(x)v(x+h)v(x)h(6)

From the properties of limits , the limit of a product is equal to the product of the limits, hence the above may be written as
f(x)=limh0v(x+h)limh0u(x+h)u(x)h+limh0u(x)limh0v(x+h)v(x)h(7)

Evaluate the individual limits in (7)
limh0v(x+h)=v(x)
limh0u(x)=u(x)
limh0u(x+h)u(x)h=u(x) , according to the definition of the derivative given in (1).
limh0v(x+h)v(x)h=v(x) , according to the definition of the derivative given in (1).

Substitute the individual limits above in (7) to obtain
f(x)=u(x)v(x)+u(x)v(x)

Hence the rule of the derivative of a product is given by (uv)=u(x)v(x)+u(x)v(x)(I)


Examples with Solutions

Example 1

Find the derivatives of
a) f(x)=xln(x)       b) g(x)=sin(x)ex

Solution
a)
Let u(x)=x and v(x)=lnx and write f(x) as the product of u and v as follows
f(x)=u(x)v(x)

Write that f(x)=(u(x)v(x))

Use the product rule in (I)
f(x)=(u(x)v(x))=u(x)v(x)+u(x)v(x)(8)

Calculte the derivatives u and v
u(x)=1 and v=1x

Substitute u,u,v,v by their expressions in (8) above to obtain f(x)=1lnx+x1x

Simplify to obtain the final answer as f(x)=lnx+1

b)
Let w(x)=sin(x) and z(x)=ex and write g(x) as the product of w and z as follows
g(x)=w(x)z(x)

Write that
g(x)=(w(x)z(x))

Use the product rule in (I)
g(x)=(w(x)z(x))=w(x)z(x)+w(x)z(x)(9)

Calculte the derivatives w and z
w(x)=cos(x) and z(x)=ex

Substitute w,w,z,z by their expressions in (9) above to obtain
g(x)=cos(x)ex+sin(x)ex

Factor ex to obtain the final answer as
g(x)=(cos(x)+sin(x))ex


Example 2

Calculate the derivative of h(x)=(2x+3)cos(x)lnx

Solution
Function h(x) is given by the product of three functions. Let u=2x+3, v=cos(x) and w=lnx and write h(x) as
h(x)=u(x)v(x)w(x)

Use the product rule in I to write
h(x)=u(x)(v(x)w(x))+u(x)(v(x)w(x))


Use the product rule in (I)m one more time to the term (v(x)w(x)) and write
h(x)=u(x)(v(x)w(x)+v(x)w(x))+u(x)(v(x)w(x))


Expand and write a form that is easy to retain

h(x)=u(x)v(x)w(x)+u(x)v(x)w(x)+u(x)v(x)w(x)(10)

Evaluate the derivatives of u,v,w u(x)=2 , v(x)=sin(x) , w=1x

Substitute u,u,v,v,w,w) by their expressions in (3) above to obtain
h(x)=2cos(x)lnx(2x+3)sin(x)lnx+(2x+3)cos(x)x


Exercises

Find the derivatives of the functions

  1. f(x)=(3x5)cos(x)
  2. g(x)=(4x+3)ex
  3. h(x)=x3sin(x)ex

Solutions to the Above Exercises

  1. f(x)=3cos(x)(3x5)sin(x)
  2. g(x)=4ex+(4x+3)ex=(4x1)ex
  3. h(x)=3x2sin(x)ex+x3cos(x)ex+x3sin(x)ex=3x2sin(x)ex+(cos(x)+sin(x))x3ex



More References and Links

  1. Quotient Rule of Differentiation
  2. Definition of Derivative
  3. Properties of Limits of Functions