The method of logarithmic differentiation in calculus uses the properties of logarithmic functions to differentiate complicated functions and functions where the usual formulas of differentiation do not apply directly. Several examples with detailed solutions are presented.
Examples with Detailed Solutions
Example 1
Find the derivative of:
\[ y = x^{\sin x} \]Solution to Example 1
We first note that there is no standard formula that can be used to differentiate this function directly. The first derivative may be calculated by taking the natural logarithm of both sides of \( y = x^{\sin x} \):
\[ \ln y = \ln \left( x^{\sin x} \right) \]Use logarithm properties to rewrite the right side:
\[ \ln y = \sin x \ln x \]We now differentiate both sides with respect to \( x \), using the chain rule on the left side and the product rule on the right side:
\[ \dfrac{y'}{y} = \cos x \ln x + \sin x \left( \dfrac{1}{x} \right) \]Multiply both sides by \( y \):
\[ y' = \left( \cos x \ln x + \dfrac{\sin x}{x} \right) y \]Substitute \( y = x^{\sin x} \):
\[ y' = \left( \cos x \ln x + \dfrac{\sin x}{x} \right) x^{\sin x} \]Example 2
Find the derivative \( y' \) of function \( y \) defined by:
\[ y = x e^{-x^2} \]Solution to Example 2
We take the natural logarithm of both sides:
\[ \ln y = \ln x + \ln e^{-x^2} \]Simplify the term \( \ln e^{-x^2} \):
\[ \ln y = \ln x - x^2 \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = \dfrac{1}{x} - 2x \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{1}{x} - 2x \right) y \]Substitute \( y = x e^{-x^2} \):
\[ y' = \left( \dfrac{1}{x} - 2x \right) x e^{-x^2} \]Simplify:
\[ y' = e^{-x^2} - 2x^2 e^{-x^2} \]Example 3
Find the derivative \( y' \) of function \( y \) given by:
\[ y = 3x^2 e^{-x} \]Solution to Example 3
We take the natural logarithm of both sides:
\[ \ln y = \ln 3 + \ln(x^2) + \ln e^{-x} \]Simplify the terms:
\[ \ln y = \ln 3 + 2 \ln x - x \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = 0 + \dfrac{2}{x} - 1 \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{2}{x} - 1 \right) y \]Substitute \( y = 3x^2 e^{-x} \):
\[ y' = \left( \dfrac{2}{x} - 1 \right) 3x^2 e^{-x} \]Rewrite as:
\[ y' = 3x(2 - x) e^{-x} \]NOTE: As an exercise, use the usual product rule of differentiation to differentiate the above function and compare results.
Example 4
Find the derivative \( y' \) of function \( y \) given by:
\[ y = (1 - x)^2 (x + 1)^4 \]Solution to Example 4
Take the natural logarithm of both sides and expand the expressions:
\[ \ln y = 2 \ln(1 - x) + 4 \ln(x + 1) \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = -2 \left( \dfrac{1}{1 - x} \right) + 4 \left( \dfrac{1}{x + 1} \right) \]Multiply all terms by \( y \):
\[ y' = \left( -2 \left( \dfrac{1}{1 - x} \right) + 4 \left( \dfrac{1}{x + 1} \right) \right) (1 - x)^2 (x + 1)^4 \]Simplify:
\[ y' = -2(1 - x)(x + 1)^3 + 4(x + 1)^3(1 - x)^2 \]NOTE: Use the usual product and chain rules to differentiate the original function directly and verify the result.
Example 5
Find the derivative \( y' \) of function \( y \) defined by:
\[ y = \dfrac{\tan x}{e^x} \]Solution to Example 5
Take the natural logarithm of both sides:
\[ \ln y = \ln(\tan x) - \ln e^x \]Simplify \( \ln e^x = x \):
\[ \ln y = \ln(\tan x) - x \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = \dfrac{\sec^2 x}{\tan x} - 1 \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{\sec^2 x}{\tan x} - 1 \right) \dfrac{\tan x}{e^x} \]Simplify:
\[ y' = \dfrac{\sec^2 x - \tan x}{e^x} \]NOTE: Use the quotient rule to differentiate this function directly and verify.
Example 6
Find the derivative \( y' \) of function \( y \) given by:
\[ y = \dfrac{(x - 2)(x + 4)}{(x + 1)(x + 5)} \]Solution to Example 6
Take the natural logarithm of both sides and expand using logarithm properties:
\[ \ln y = \ln(x - 2) + \ln(x + 4) - \ln(x + 1) - \ln(x + 5) \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = \dfrac{1}{x - 2} + \dfrac{1}{x + 4} - \dfrac{1}{x + 1} - \dfrac{1}{x + 5} \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{1}{x - 2} + \dfrac{1}{x + 4} - \dfrac{1}{x + 1} - \dfrac{1}{x + 5} \right) \dfrac{(x - 2)(x + 4)}{(x + 1)(x + 5)} \]Simplify:
\[ y' = \dfrac{2(2x^2 + 13x + 29)}{(x + 1)^2 (x + 5)^2} \]NOTE: Use the quotient rule directly to compare results.
Example 7
Use logarithmic differentiation to prove the product rule \( y = uv \), where \( u \) and \( v \) are functions of \( x \).
Solution to Example 7
Take the natural logarithm of both sides and expand:
\[ \ln y = \ln u + \ln v \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = \dfrac{u'}{u} + \dfrac{v'}{v} \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{u'}{u} + \dfrac{v'}{v} \right) y = \left( \dfrac{u'}{u} + \dfrac{v'}{v} \right) uv \] \[ y' = u'v + v'u \]NOTE: The result obtained is the well-known product rule of differentiation.
Example 8
Use logarithmic differentiation to prove the quotient rule \( y = \dfrac{u}{v} \), where \( u \) and \( v \) are functions of \( x \).
Solution to Example 8
Take the natural logarithm of both sides and expand using logarithm properties:
\[ \ln y = \ln u - \ln v \]Differentiate both sides with respect to \( x \):
\[ \dfrac{y'}{y} = \dfrac{u'}{u} - \dfrac{v'}{v} \]Multiply all terms by \( y \):
\[ y' = \left( \dfrac{u'}{u} - \dfrac{v'}{v} \right) y = \left( \dfrac{u'}{u} - \dfrac{v'}{v} \right) \dfrac{u}{v} \] \[ y' = \dfrac{u'v - v'u}{v^2} \]NOTE: The result obtained is the well-known quotient rule of differentiation.