# Logarithmic Differentiation Method

The method of logarithmic differentiation , calculus, uses the properties of logarithmic functions to differentiate complicated functions and functions where the usual formulas of Differentiation do not apply. Several examples with detailed solutions are presented.

## Example 1

$y = x^{ \sin x }$

## Solution to Example 1

• We first note that there is no formula that can be used to differentiate directly this function. The first derivative may be calculated by first 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 above equation as follows
$$\ln y = \sin x \ln x$$
• We now differentiate both sides with respect to $$x$$, using the chain rule on the left side and the product rule formula for differentiation on the right side.
$$\dfrac{y'}{y} = \cos x \ln x + \sin x \left( \dfrac{1}{x} \right)$$
• Multiply both sides by $$y$$ to obtain
$$y' = \left( \cos x \ln x + \dfrac{\sin x}{x} \right) y$$
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 logarithms 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 $$y$$ and simplify
$$y' = \left( \dfrac{1}{x} - 2x \right) y$$
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 $y = 3 x^{2} e^{ -x }$

## Solution to Example 3

• We take the logarithms of both sides of the given function
$$\ln y = \ln 3 + \ln x^{2} + \ln e^{ -x }$$
• Simplify the term $$\ln e^{ -x }$$
$$\ln y = \ln 3 + 2 \ln x - x$$
• 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 $$y = 3 x^{2} e^{ -x }$$
$$y' = \left( \dfrac{2}{x} - 1 \right) 3 x^{2} e^{ -x }$$
Rewrite as
$$y' = 3x \left( 2 - x \right) e^{ -x }$$
NOTE: As an exercise, use the usual formula 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 logarithms of both sides and expand expressions obtained
$$\ln y = 2 \ln (1 - x) + 4 \ln (x + 1)$$
• 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 $$y$$ and simplify
$$y' = \left( -2 \left( \dfrac{1}{1 - x} \right) + 4 \left( \dfrac{1}{x + 1} \right) \right) y$$
$$y' = \left( -2 \left( \dfrac{1}{1 - x} \right) + 4 \left( \dfrac{1}{x + 1} \right) \right) (1 - x)^{2} (x + 1)^{4}$$
$$y' = -2\left(1-x\right)\left(x+1\right)^4+4\left(x+1\right)^3\left(1-x\right)^2$$
NOTE: Use the usual formula of differentiation to differentiate the above function and compare results.

## Example 5

Find the derivative $$y'$$ of function $$y$$ defined by
$y = \dfrac{ \tan x }{ e^{ x } }$

## Solution to Example 5

• Take the logarithms of both sides
$$\ln y = \ln (\tan x) - \ln e^{ x }$$
• Simplify $$\ln e^{ x }$$.
$$\ln y = \ln (\tan x) - x$$
• Differentiate both sides with respect to $$x$$
$$\dfrac{y'}{y} = \dfrac{\sec^{2} x}{\tan x} - 1$$
• Multiply all terms by $$y$$
$$y' = \left( \dfrac{\sec^{2} x}{\tan x} - 1 \right) y$$
$$y' = \left( \dfrac{\sec^{2} x}{\tan x} - 1 \right) \dfrac{\tan x}{e^{ x }}$$
Simplify
$$y' = \dfrac{\sec^{2} x - \tan x}{e^{ x }}$$
NOTE: Use the usual formula of differentiation to differentiate the above function and compare results.

## 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 logarithms of both sides and expand the expressions obtained
$$\ln y = \ln (x - 2) + \ln (x + 4) - \ln (x + 1) - \ln (x + 5)$$
• 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 $$y$$ and simplify to obtain
$$y' = \left( \dfrac{1}{(x - 2)} + \dfrac{1}{(x + 4)} - \dfrac{1}{(x + 1)} - \dfrac{1}{(x + 5)} \right) y$$
$$y' = \left( \dfrac{1}{(x - 2)} + \dfrac{1}{(x + 4)} - \dfrac{1}{(x + 1)} - \dfrac{1}{(x + 5)} \right) \dfrac{(x - 2)(x + 4)}{(x + 1)(x + 5)}$$
$$y' = \dfrac{2(2x^{2} + 13x + 29)}{(x + 1)^{2}(x + 5)^{2}}$$
NOTE: Use the usual formula of differentiation to differentiate the above function and compare results.

## Example 7

Use the method of taking the logarithms to find $$y'$$ if $$y = uv$$, where $$u$$ and $$v$$ are functions of $$x$$.

## Solution to Example 7

• Take the logarithms of both sides and expand the expressions obtained
$$\ln y = \ln u + \ln v$$
• Differentiate both sides with respect to $$x$$
$$\dfrac{y'}{y} = \dfrac{u'}{u} + \dfrac{v'}{v}$$
• Multiply all terms by $$y$$ and simplify to obtain $$y' = \left( \dfrac{u'}{u} + \dfrac{v'}{v} \right) y$$
$$y' = \left( \dfrac{u'}{u} + \dfrac{v'}{v} \right) uv$$
$$y' = u'v + v'u$$
NOTE: The result obtained is the well known product rule of differentiation.

## Example 8

Use the method of taking the logarithms to find $$y'$$ if $$y = \dfrac{u}{v}$$, where $$u$$ and $$v$$ are functions of $$x$$.

## Solution to Example 8

• Take the logarithms of both sides and expand the expressions obtained using the logarithm properties
$$\ln y = \ln u - \ln v$$
• Differentiate both sides with respect to $$x$$ using the differentiation rule of the logarithm of a function
$$\dfrac{y'}{y} = \dfrac{u'}{u} - \dfrac{v'}{v}$$
• Multiply all terms by $$y$$ and simplify to obtain
$$y' = \left( \dfrac{u'}{u} - \dfrac{v'}{v} \right) y$$
$$y' = \left( \dfrac{u'}{u} - \dfrac{v'}{v} \right) \dfrac{u}{v}$$
$$y' = \dfrac{u'v - v'u}{v^{2}}$$
NOTE: The result obtained is the well known quotient rule of differentiation of functions.

### More References and links

differentiation and derivatives