The steps to prove the product rule of differentiation are presented along with examples, exercises and solutions.
The derivative f′(x) of the function f(x) is defined
as
f′(x)=limh→0f(x+h)−f(x)h(1)
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)=limh→0u(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)=limh→0u(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)=limh→0(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)=limh→0u(x+h)v(x+h)−u(x)v(x+h)h+limh→0(u(x)v(x+h)−u(x)v(x)h)(5)
Use factoring to rewrite the above as
f′(x)=limh→0v(x+h)u(x+h)−u(x)h+limh→0u(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)=limh→0v(x+h)limh→0u(x+h)−u(x)h+limh→0u(x)limh→0v(x+h)−v(x)h(7)
Evaluate the individual limits in (7)
limh→0v(x+h)=v(x)
limh→0u(x)=u(x)
limh→0u(x+h)−u(x)h=u′(x) , according to the definition of the derivative given in (1).
limh→0v(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)
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)=1⋅lnx+x⋅1x
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
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
Find the derivatives of the functions