The steps to prove the product rule of differentiation are presented along with examples, exercises, and solutions.
Derivative of the Product of Two Functions (Proof)
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:
\[ f(x) = u(x)v(x) \]Using the definition of the derivative in (1), the derivative of \( f(x) = u(x)v(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x)}{h} \qquad (2) \]Subtracting and adding the same quantity (\( u(x)v(x+h) \)) in the numerator of (2) does not change its value:
\[ f'(x) = \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x) - u(x)v(x+h) + u(x)v(x+h)}{h} \qquad (3) \]Split the fraction into two parts:
\[ f'(x) = \lim_{h \to 0} \left( \dfrac{u(x+h)v(x+h) - u(x)v(x+h)}{h} + \dfrac{u(x)v(x+h) - u(x)v(x)}{h} \right) \qquad (4) \]From the properties of limits, the limit of a sum is equal to the sum of the limits:
\[ f'(x) = \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x+h)}{h} + \lim_{h \to 0} \dfrac{u(x)v(x+h) - u(x)v(x)}{h} \qquad (5) \]Use factoring to rewrite the terms:
\[ f'(x) = \lim_{h \to 0} \left[ v(x+h) \dfrac{u(x+h) - u(x)}{h} \right] + \lim_{h \to 0} \left[ u(x) \dfrac{v(x+h) - v(x)}{h} \right] \qquad (6) \]Since the limit of a product is the product of the limits:
\[ f'(x) = \left( \lim_{h \to 0} v(x+h) \right) \left( \lim_{h \to 0} \dfrac{u(x+h) - u(x)}{h} \right) + \left( \lim_{h \to 0} u(x) \right) \left( \lim_{h \to 0} \dfrac{v(x+h) - v(x)}{h} \right) \qquad (7) \]Evaluating the individual limits:
- \( \lim_{h \to 0} v(x+h) = v(x) \)
- \( \lim_{h \to 0} u(x) = u(x) \)
- \( \lim_{h \to 0} \dfrac{u(x+h) - u(x)}{h} = u'(x) \) (by definition 1)
- \( \lim_{h \to 0} \dfrac{v(x+h) - v(x)}{h} = v'(x) \) (by definition 1)
Substitute these limits back into (7) to obtain:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]Examples with Solutions
Example 1
Find the derivatives of:
a) \( f(x) = x \ln(x) \) b) \( g(x) = \sin(x)e^x \)
Solution to Example 1
a) Let \( u(x) = x \) and \( v(x) = \ln x \). Write \( f(x) = u(x)v(x) \).
Using the product rule formula (I):
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]Since \( u'(x) = 1 \) and \( v'(x) = \dfrac{1}{x} \):
\[ f'(x) = 1 \cdot \ln x + x \cdot \dfrac{1}{x} = \ln x + 1 \]b) Let \( w(x) = \sin(x) \) and \( z(x) = e^x \). Write \( g(x) = w(x)z(x) \).
Using the product rule formula (I):
\[ g'(x) = w'(x)z(x) + w(x)z'(x) \]Since \( w'(x) = \cos(x) \) and \( z'(x) = e^x \):
\[ g'(x) = \cos(x)e^x + \sin(x)e^x = (\cos x + \sin x)e^x \]Example 2
Calculate the derivative of:
\[ h(x) = (2x + 3)\cos(x)\ln x \]Solution to Example 2
Function \( h(x) \) is the product of three functions. Let \( u = 2x + 3 \), \( v = \cos(x) \), and \( w = \ln x \), so \( h(x) = u(x)v(x)w(x) \).
Applying the product rule for three functions:
\[ h'(x) = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x) \]Evaluate the individual derivatives:
\[ u'(x) = 2, \quad v'(x) = -\sin(x), \quad w'(x) = \dfrac{1}{x} \]Substitute these expressions into the formula:
\[ h'(x) = 2\cos(x)\ln x - (2x + 3)\sin(x)\ln x + \dfrac{(2x + 3)\cos(x)}{x} \]Exercises
Find the derivatives of the functions:
- \( f(x) = (3x - 5)\cos(x) \)
- \( g(x) = (-4x + 3)e^x \)
- \( h(x) = x^3 \sin(x)e^x \)
Solutions to the Above Exercises
- \( f'(x) = 3\cos(x) - (3x - 5)\sin(x) \)
- \( g'(x) = -4e^x + (-4x + 3)e^x = (-4x - 1)e^x \)
- \( h'(x) = 3x^2\sin(x)e^x + x^3\cos(x)e^x + x^3\sin(x)e^x = 3x^2\sin(x)e^x + (\cos(x) + \sin(x))x^3e^x \)