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

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

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
\[ (u v)' = u'(x) v(x) + u(x) v'(x) \qquad (I) \]

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

\[ \quad g'(x) = \left( cos(x) + sin(x) \right) \; e^x \]

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

\[ \quad h'(x) = 2 \cos(x) \ln x - (2x+3) \sin(x) \ln x + \dfrac{(2x+3) \cos(x)}{x} \]

Find the derivatives of the functions

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

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