Logarithmic Differentiation
Logarithmic differentiation is used to differentiate large functions, with the use of logarithms and chain rule of differentiation. The logarithmic differentiation of a function f(x) is f'(x)/f(x)· \(\dfrac{d}{dx}.log f(x) = \dfrac{f'(x)}{f(x)}\)· Also the use of logarithms transforms the product of functions into sum of functions, and the division of functions into difference of functions.
Exponential functions or functions with a lot of subfunctions can be easily differentiated using logarithmic differentiation. Let us learn more about the applications of logarithmic differentiation with examples.
1.  What is Logarithmic Differentiation? 
2.  Applications Of Log Differentiation 
3.  Examples on Logarithmic Differentiation 
4.  Practice Questions 
5.  FAQs on Logarithmic Differentiation 
What is Logarithmic Differentiation?
Logarithmic differentiation is based on the logarithm properties and the chain rule of differentiation and is mainly used to differentiate functions of the form f(x)^{g(x)·} It helps in easily performing the differentiation in simple and quick steps. The functions which are complex and cannot be algebraically solved and differentiated can be differentiated using logarithmic differentiation.
Logarithmic Differentiation Formula
The logarithmic differentiation of a function f(x) is equal to the differentiation of the function, divided by the function. Here is the formula that is used mainly in logarithmic differentiation.
\(\frac{d}{dx} \log f(x) = \dfrac{f'(x)}{f(x)}\)
Logarithmic differentiation is used if the function is made of a number of subfunctions, with a product between the functions, the division between the functions, an exponential relationship between the functions, or a function raised to another function. Logarithms help in transforming the product of functions into the sum of functions, and the division of functions into differences of functions. Further after breaking the function using logarithms, it can be easily differentiated as a normal function, using the chain rule of differentiation. The chain rule of differentiation first differentiated the function involving logarithms and then differentiates the function independently. d/dx log f(x) = 1/f(x) d/dx f(x)·
\(\dfrac{d}{dx}.log f(x) = \dfrac{1}{f(x)} \dfrac{d}{dx} f(x)\)
The following set of logarithmic properties help in easily simplifying the functions and in performing the differentiation process.
 log AB = log A + log B
 log A/B = log A  log B
 log A^{B} = B log A
 log_{B}A = (log A) / (log B)
Applications of Log Differentiation
The log differentiation has applications for the product of functions, for the division of two functions, and for exponential functions. Let us look into each of these applications of logarithmic differentiation.
Product of Functions
For the product of two or more functions, the application of logarithms transforms the product into a sum of functions and facilitates easy differentiation of the function. Let the function f(x), be the product of two subfunctions g(x), and h(x) respectively, and we can apply logarithmic followed by differentiation of the functions.
f(x) = g(x) · h(x)
Let us apply logarithms on both sides of the above equation representing the product of functions.
log f(x) = log (g(x) · h(x))
log f(x) = log g(x) + log h(x)
Let us now differentiate on both sides.
d/dx log f(x) = d/dx log g(x) + d/dx log h(x)
f'(x)/f(x) = g'(x)/g(x) + h'(x)/h(x)
f'(x) = f(x) [g'(x)/g(x) + h'(x)/h(x)]
f'(x) = f(x) [h(x)·g'(x) + g(x)·h'(x)] / g(x)·h(x)
f'(x) = g(x)·h(x) [h(x)·g'(x) + g(x)·h'(x)] / g(x)·h(x)
f'(x) = h(x)·g'(x) + g(x)·h'(x)
This differentiation of the product of two functions, involving logarithmic differentiation is called as Leibniz rule. The above rule is known by the name "Product rule".
Division of Functions
The differentiation of the division of one function with another function also called the quotient of functions is obtained by the process of logarithmic differentiation. The application of logarithms to the division of one function with another function transforms it into a difference in the logarithms of each of the two functions. Let us consider a function f(x), which is equal to the quotient of the two functions g(x) and h(x).
f(x) = g(x)/h(x)
Let us apply logarithms on both sides of the above equal representing the quotient of the two functions.
log f(x) = log g(x)/h(x)
log f(x) = log g(x)  log h(x)
Further we can apply differentiation to the above logarithmic equation.
d/dx log f(x) = d/dx log g(x)  d/dx log h(x)
f'(x)/f(x) = g'(x)/g(x)  h'(x)/h(x)
f'(x) = f(x)[g'(x)/g(x)  h'(x)/h(x)]
f'(x) = f(x) [g'(x)·h(x)  g(x)·h'(x)]/g(x)·h(x)
f'(x) = g(x)/h(x) [g'(x)·h(x)  g(x)·h'(x)]/g(x)·h(x)
f'(x) = [g'(x)·h(x)  g(x)·h'(x)]/h^{2}(x)
The above rule is known as "Quotient rule".
Exponential Functions
Here we consider two functions such that one is an exponent of another function, and the application of logarithms to this function transforms it into a product of one function and the logarithm of the other function. Let us take a function f(x), which is equal to the exponent of g(x) to h(x).
f(x) = g(x)^{h(x)}
Here we apply logarithms on both sides.
log f(x) = log g(x)^{h(x)}
log f(x) = h(x) · log g(x)
Further we apply differentiation on both sides.
d/dx log f(x) = d/dx [h(x) · log g(x)]
f'(x)/f(x) = h(x)· d/dx log g(x) + log g(x)· d/dx h(x)
f'(x)/f(x) = h(x)· g'(x)/g(x) + log g(x)· h'(x)
f'(x)/f(x) = [h(x)·g'(x) + g(x)·h'(x)·log g(x)]/g(x)
f'(x) = f(x) [h(x)·g'(x) + g(x)·h'(x)·log g(x)]/g(x)
f'(x) = g(x)^{h(x)·}[h(x)·g'(x) + g(x)·h'(x)·log g(x)]/g(x)
f'(x) = g(x)^{h(x)1} [h(x)·g'(x) + g(x)·h'(x)·log g(x)]
Important Notes on Logarithmic Differentiation:
 Logarithmic differentiation is mandatory to be used when we need to find the derivative of a function that is of the form h(x) = f(x)^{g(x)}.
 Here, both h(x) and f(x) need to be positive in the given domain for making the process meaningful.
ā Related Topics:
The following topics help for a better understanding of logarithmic differentiation.
Examples on Logarithmic Differentiation

Example 1: Find the differentiation of x^{tan x} using the concept of logarithmic differentiation.
Solution:
The given expression for differentiation is y = x^{tan x}.
Let us apply logarithms on both sides of the expression.
log y = log x^{tan x}
log y = tan x log x.
Apply differentiation on both sides.
d/dx log y = d/dx (tan x log x)
1/y · dy/dx = log x (d/dx tan x) + tan x (d/dx log x)
1/y · dy/dx = log x(sec^{2}x) + tan x(1/x)
1/y · dy/dx = sec^{2}x log x + tan x/x
1/y · dy/dx = (x sec^{2}x log x + tan x)/x
dy/dx = y(x sec^{2}x log x + tan x)/x
dy/dx = x^{tan x}(x sec^{2}x log x + tan x)/x
Answer: Therefore, the differentiation of x^{tan x} is x^{tan x}(x sec^{2}x log x + tan x)/x.

Example 2: What is the differentiation of (x + 1)(2x + 3)(5x + 4)?
Solution:
The given expression is y = (x + 1)(2x + 3)(5x + 4)
Here we can use the concept of logarithmic differentiation.
Let us first apply logarithms on both sides.
log y = log (x + 1)(2x + 3)(5x + 4)
log y = log (x + 1) + log (2x + 3) + log (5x + 4)
d/dx log y = d/dx log (x + 1) + d/dx log (2x + 3) + d/dx log (5x + 4)
1/y · dy/dx = 1/(x + 1) · d/dx(x + 1) + 1/(2x + 3) · d/dx(2x + 3) + 1/(5x + 4) · d/dx(5x + 4)
1/y · dy/dx = 1/(x + 1)·1 + 1/(2x + 3)·2 + 1/(5x + 4)·5
1/y · dy/dx = 1/(x + 1) + 2/(2x + 3) + 5/(5x + 4)
dy/dx = y [1/(x + 1) + 2/(2x + 3) + 5/(5x + 4)]
dy/dx = (x + 1)(2x + 3)(5x + 4) [1/(x + 1) + 2/(2x + 3) + 5/(5x + 4)]
dy/dx = (2x + 3)(5x + 4) + 2(x + 1)(5x + 4) + 5(x + 1)(2x + 3)
dy/dx = (5x + 4) [(2x + 3) + 2(x + 1)] + 5(x + 1)(2x + 3)
dy/x = (5x + 4)(2x + 3 + 2x + 2) + 5(x + 1)(2x + 3)
dy/dx = (5x + 4)(4x + 5) + 5(x + 1)(2x + 3)
dy/dx = 20x^{2} + 41x + 20 + 10x^{2} + 25x + 15
dy/dx = 30x^{2} + 66x + 35Answer: Therefore, the differentiation of the function (x + 1)(2x + 3)(5x + 4) is 30x^{2} + 66x + 35.

Example 3: Find the derivative of e\(^{x^3}\) using logarithmic differentiation. Also, verify the result by any other method.
Solution:
Let y = e\(^{x^3}\)
Derivative by Logarithmic Differentiation:
Taking ln on both sides,
ln y = ln e\(^{x^3}\)
ln y = x^{3} ln e
ln y = x^{3} (āµ ln e = 1)
Taking derivative with respect to x on both sides,
1/y dy/dx = 3x^{2}
dy/dx = y 3x^{2} = 3x^{2} e\(^{x^3}\)
Derivative by Chain Rule:
We know that the derivative of e to the power of x is itself. Also, by chain rule,
dy/dx = e\(^{x^3}\) d/dx (x^{3}) = e\(^{x^3}\) (3x^{2}) = 3x^{2} e\(^{x^3}\)
Answer: dy/dx = 3x^{2} e\(^{x^3}\) and the answer is verified by alternative method.
FAQs on Logarithmic Differentiation
What is the Formula of Logarithmic Differentiation?
The logarithmic differentiation of a function f(x) is equal to the differentiation of the function divided by the function. i.e., d/dx (log f(x)) = f '(x)/f(x). The logarithmic differentiation of a function takes the advantage of the logarithm concepts and the chain rule of differentiation. Further, it can be used for the differentiation of one function raised to another function.
How To Use Log Differentiation?
The log differentiation can be used along with logarithm formulas and with the concept of chain rule of differentiation. Functions that are a product of multiple subfunctions, or when one function is divided by another function, or if a function is an exponent of another function, etc can be differentiated with the help of logarithmic differentiation.
When Do We Need Logarithmic Differentiation?
Logarithmic differentiation is needed when there is a complex function that is a product of numerous subfunctions or if one function is a quotient of another function. Often logarithmic differentiation is needed when one function is an exponent of another function f(x)^{g(x)}.
What Are The Advantages Of Logarithmic Differentiation?
The logarithmic differentiation has the advantage of performing the differentiation of a complicated function in a few simple steps. The application of logarithms to the function transforms multiplication intoaddition and division into subtraction. Here the use of logarithm concepts makes the process of differentiation easier.
What Are Log Differentiation Examples?
We use log differentiation to find the derivatives of functions with exponents as functions like tan x^{cos x}, difficult products like (x + 1)^{2} (2x + 3)^{3}, difficult quotients like √ [ ((x + 1) (x  2)) / (2x + 1) (3x  2) ].
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