Product Rule of Differentiation with Examples

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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
\[ f'(x) = \lim_{h \to 0} \; \dfrac{f(x+h) - f(x)}{h} \qquad (1) \]


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
\[ \quad f(x) = u(x) v(x) \]
Using the definition of the derivative in \( (1) \), the derivative of \( f(x) = u(x) v(x) \) is given by

\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) - u(x) v(x)}{h} \qquad (2) \)

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

\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) \color{red}{- u(x) v(x+h)} - u(x) v(x) \color{red}{+ u(x) v(x+h)} }{h} \qquad (3) \)

Split \( (3) \) and rewrite it as follows
\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \left( \; \dfrac{u(x+h) v(x+h) - u(x) v(x+h)}{h} + \dfrac{u(x) v(x+h) - u(x) v(x) }{h} \right) \qquad (4) \)

From the properties of limits , the limit of a sum is equal to the sum of the limits, hence
\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) - u(x) v(x+h)}{h} + \lim_{h \to 0} \left ( \dfrac{u(x) v(x+h) - u(x) v(x) }{h} \right) \qquad (5) \)

Use factoring to rewrite the above as
\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \; v(x+h) \dfrac{u(x+h) - u(x) }{h} + \lim_{h \to 0} u(x) \dfrac{ v(x+h) - v(x) }{h} \qquad (6) \)

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
\( \qquad \displaystyle f'(x) = \lim_{h \to 0} \; v(x+h) \lim_{h \to 0} \; \dfrac{u(x+h) - u(x) }{h} + \lim_{h \to 0} u(x) \lim_{h \to 0} \; \dfrac{ v(x+h) - v(x) }{h} \qquad (7) \)

Evaluate the individual limits in \( (7) \)
\( \qquad \displaystyle \lim_{h \to 0} \; v(x+h) = v(x) \)
\( \qquad \displaystyle \lim_{h \to 0} u(x) = u(x) \)
\( \qquad \displaystyle \lim_{h \to 0} \; \dfrac{u(x+h) - u(x) }{h} = u'(x) \) , according to the definition of the derivative given in \( (1) \).
\( \qquad \displaystyle \lim_{h \to 0} \; \dfrac{ v(x+h) - v(x) }{h} = v'(x) \) , according to the definition of the derivative given in \( (1) \).

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

Hence the rule of the derivative of a product is given by \[ (u v)' = u'(x) v(x) + u(x) v'(x) \qquad (I) \]


Examples with Solutions

Example 1

Find the derivatives of
a) \( \quad f(x) = x \; \ln(x) \)       b) \( \quad g(x) = \sin(x) e^x \)

Solution
a)
Let \( u(x) = x \) and \( v(x) = \ln x \) and write \( f(x) \) as the product of \( u \) and \( v \) as follows
\( \quad f(x) = u(x) \; v(x) \)

Write that \( \quad f'(x) = (u(x) \; v(x))' \)

Use the product rule in \( (I) \)
\( \quad f'(x) = (u(x) \; v(x))' = u'(x) v(x) + u(x) v'(x) \quad (8) \)

Calculte the derivatives \( u' \) and \( v' \)
\( \quad u'(x) = 1 \) and \( v' = \dfrac{1}{x} \)

Substitute \( u, u', v, v' \) by their expressions in \( (8) \) above to obtain \( \quad f'(x) = 1 \cdot \ln x + x \cdot \dfrac{1}{x} \)

Simplify to obtain the final answer as \[ \quad f'(x) = \ln x + 1 \]

b)
Let \( w(x) = \sin(x) \) and \( z(x) = e^x \) and write \( g(x) \) as the product of \( w \) and \( z \) as follows
\( \quad g(x) = w(x) \; z(x) \)

Write that
\( \quad g'(x) = ( w(x) \; z(x) )' \)

Use the product rule in \( (I) \)
\( \quad g'(x) = (w(x) \; z(x))' = w'(x) z(x) + w(x) z'(x) \quad (9) \)

Calculte the derivatives \( w' \) and \( z' \)
\( \quad w'(x) = cos(x) \) and \( z'(x) = e^x \)

Substitute \( w, w', z, z' \) by their expressions in \( (9) \) above to obtain
\( \quad g'(x) = cos(x) \cdot e^x + sin(x) \cdot e^x \)

Factor \( e^x \) to obtain the final answer as
\[ \quad g'(x) = \left( cos(x) + sin(x) \right) \; e^x \]


Example 2

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

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

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


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


Expand and write a form that is easy to retain

\( \quad h'(x) = \color{red}{u'(x)} v(x) w(x) + u(x) \color{red}{v'(x)} w(x) + u(x) v(x) \color{red}{w'(x)} \quad (10) \)

Evaluate the derivatives of \( u, v , w \) \( u'(x) = 2 \) , \( v'(x) = - \sin(x) \) , \( w' = \dfrac{1}{x} \)

Substitute \( u, u', v, v', w, w') \) by their expressions in \( (3) \) above to obtain
\[ \quad h'(x) = 2 \cos(x) \ln x - (2x+3) \sin(x) \ln x + \dfrac{(2x+3) \cos(x)}{x} \]


Exercises

Find the derivatives of the functions

  1. \( \quad f(x) = (3x-5) \cos(x) \)
  2. \( \quad g(x) = (-4x+3) e^x \)
  3. \( \quad h(x) = x^3 \sin(x) e^x \)

Solutions to the Above Exercises

  1. \( \quad f'(x) = 3\cos (x) - (3x-5) \sin (x) \)
  2. \( \quad g'(x) = -4e^x + (-4x+3)e^x = (-4 x - 1) e^x \)
  3. \( \quad h'(x) = 3x^2\sin (x)e^x + x^3 \cos (x)e^x + x^3 \sin(x) e^x = 3x^2\sin (x)e^x + (\cos (x) + \sin(x) ) x^3 e^x\)



More References and Links

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

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