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

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