The steps to prove the quotient rule of differentiation from the product rule of differentiation are presented along with examples, exercises, and solutions.
Derivative of the Quotient of Two Functions (Proof)
Let function \( f(x) \) be given by the quotient of two functions \( u(x) \) and \( v(x) \), written as:
\[ f(x) = \dfrac{u(x)}{v(x)} \qquad (1) \]Multiply both sides of the above by \( v(x) \) and simplify:
\[ f(x)v(x) = u(x) \]Take the derivative of both sides with respect to \( x \):
\[ (f(x)v(x))' = u'(x) \]Apply the product rule of differentiation to the left side:
\[ f'(x)v(x) + f(x)v'(x) = u'(x) \]Solve the equation for \( f'(x) \):
\[ f'(x) = \dfrac{u'(x) - f(x)v'(x)}{v(x)} \]Substitute \( f(x) \) by \( \dfrac{u(x)}{v(x)} \) as given in (1):
\[ f'(x) = \dfrac{u'(x) - \dfrac{u(x)}{v(x)}v'(x)}{v(x)} \]Rewrite the numerator with a common denominator and simplify:
\[ f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]Examples with Solutions
Example 1
Find the derivatives of:
a) \( f(x) = \dfrac{x}{\ln x} \) b) \( g(x) = \dfrac{\sin(x)}{\cos(x)} \)
Solution to Example 1
a) Let \( u(x) = x \) and \( v(x) = \ln x \). Write \( f(x) = \dfrac{u}{v} \).
Using the quotient rule (I):
\[ f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \qquad (2) \]Since \( u'(x) = 1 \) and \( v'(x) = \dfrac{1}{x} \):
\[ f'(x) = \dfrac{1 \cdot \ln x - x \cdot \dfrac{1}{x}}{(\ln x)^2} = \dfrac{\ln x - 1}{(\ln x)^2} \]b) Let \( w(x) = \sin x \) and \( z(x) = \cos x \). Write \( g(x) = \dfrac{w}{z} \).
Using the quotient rule (I):
\[ g'(x) = \dfrac{w'(x)z(x) - w(x)z'(x)}{(z(x))^2} \qquad (3) \]Since \( w'(x) = \cos x \) and \( z'(x) = -\sin x \):
\[ g'(x) = \dfrac{\cos x \cos x - \sin x(-\sin x)}{(\cos x)^2} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} \]Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ g'(x) = \dfrac{1}{\cos^2 x} = \sec^2 x \]Example 2
Calculate the derivative of:
\[ h(x) = \dfrac{x \ln x}{\sin x \, e^x} \]Solution to Example 2
Let \( u(x) = x \ln x \) and \( v(x) = \sin x \, e^x \). Write \( h(x) = \dfrac{u(x)}{v(x)} \).
Using the quotient rule (I):
\[ h'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \qquad (4) \]Calculate \( u'(x) \) using the product rule:
\[ u'(x) = (x)'\ln x + x(\ln x)' = 1 \cdot \ln x + x \cdot \dfrac{1}{x} = \ln x + 1 \]Calculate \( v'(x) \) using the product rule:
\[ v'(x) = (\sin x)'e^x + \sin x(e^x)' = \cos x \, e^x + \sin x \, e^x = (\cos x + \sin x)e^x \]Substitute into (4):
\[ h'(x) = \dfrac{(\ln x + 1)(\sin x \, e^x) - (x \ln x)((\cos x + \sin x)e^x)}{(\sin x \, e^x)^2} \]Factor out and cancel \( e^x \):
\[ h'(x) = \dfrac{(\ln x + 1)\sin x - x \ln x(\cos x + \sin x)}{\sin^2 x \, e^x} \]Exercises
Find the derivatives of the functions:
- \( f(x) = \dfrac{\cos(x)}{\sin x} \)
- \( g(x) = \dfrac{-2x + 4}{e^x} \)
- \( h(x) = \dfrac{2x \, e^x}{x \sin x} \)
Solutions to the Above Exercises
- \( f'(x) = \dfrac{-\sin x \sin x - \cos x \cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\csc^2 x \)
- \( g'(x) = \dfrac{-2e^x - (-2x + 4)e^x}{(e^x)^2} = -\dfrac{2(-x + 3)}{e^x} \)
- \( h'(x) = \dfrac{(2(1 + x)e^x)(x \sin x) - (2x \, e^x)(\sin x + x \cos x)}{(x \sin x)^2} = 2(\csc x - \cot x \csc x)e^x \)