Logarithmic Derivatives occur when we differentiate Logarithmic Functions.
The derivative of the Natural Logarithm can be expressed algebraically as:
\(\cfrac{d}{dx}\cdot \text{ln}(x) = \cfrac{1}{x}\)
This applies to functions where \(x > 0\).
The derivative of the Logarithm can be expressed algebraically as:
\(\cfrac{d}{dx}\cdot [a\cdot \text{log}_b(x)] = \cfrac{\textcolor{red}{a}}{\textcolor{blue}{x\cdot \text{ln}(b)}}\)
- \(\textcolor{red}{a}\) represents the logarithm's coefficient
- \(\textcolor{blue}{b}\) represents the logarithm's base
A Common Logarithm is a logarithm with a base of \(10\). Its derivative can be expressed as:
\(\cfrac{d}{dx}\cdot \text{log}(x) = \cfrac{1}{x\cdot \text{ln}(10)}\)
NOTE: The default value of \(a\) is \(1\).
Example
Find the derivative of each logarithm.
- \(y = \text{log}_2x\)
- \(y = \sqrt{\text{ln}x}\)
i. We can use the logarithm formula in order to determine the derivative:
\(\cfrac{d}{dx}\cdot [a\cdot \text{log}_b(x)] = \cfrac{a}{x\cdot \text{ln}(b)}\)
In this equation, we can identify that \(a = 1\) and \(b = 2\). As a result, we can express the derivative as:
\(y' = \cfrac{\textcolor{red}{1}}{\textcolor{blue}{x\cdot \text{ln}(2)}}\)
Therefore, we can determine that \(y' = \cfrac{1}{x\cdot \text{ln}(2)}\).
ii. We can use Chain Rule in order to determine the derivative.
First, we can rewrite the equation as such:
\(y' = (\text{ln}x)^{\frac{1}{2}}\)
Next, we can find the derivative of the outer function, \(f'(g(x))\):
\(f'(g(x)) = \cfrac{1}{2}(\text{ln}x)^{-\frac{1}{2}}\)
\(\textcolor{red}{f'(g(x)) = \cfrac{1}{2(\text{ln}x)^{\frac{1}{2}}}}\)
Then, we can find the derivative of the inner function, \(g(x)\), by using Natural Logarithm Rule:
\(\cfrac{d}{dx}\cdot \text{ln}(x) = \cfrac{1}{x}\)
\(\textcolor{blue}{g'(x) = \cfrac{1}{x}}\)
Finally, we can implement Chain Rule to determine the derivative:
\(h'(x) = \textcolor{red}{f'(g(x))}\textcolor{blue}{g'(x)}\)
\(y' = \textcolor{red}{\cfrac{1}{2}(\text{ln}x)^{-\frac{1}{2}}}\cdot \textcolor{blue}{\cfrac{1}{x}}\)
\(y' = \cfrac{1}{2x\cdot (\text{ln}x)^{\frac{1}{2}}}\)
\(y' = \cfrac{1}{2x\cdot \sqrt{\text{ln}x}}\)
Therefore, we can determine that \(y' = \cfrac{1}{2x\cdot \sqrt{\text{ln}x}}\).
Find the derivative of each logarithm.
\(y = -(\text{log}x)\)
Show Answer
We can use the logarithm formula in order to determine the derivative:
\(\cfrac{d}{dx}\cdot [a\cdot \text{log}_b(x)] = \cfrac{a}{x\cdot \text{ln}(b)}\)
In this function, we can identify that \(a = -1\) and \(b = 10\) (since it's the default value). As a result, we can express the derivative as:
\(y' = \cfrac{\textcolor{red}{-1}}{\textcolor{blue}{x\cdot \text{ln}(10)}}\)
Therefore, we can determine that \(y' = \cfrac{-1}{x\cdot \text{ln}(10)}\).
\(y = -3\text{log}_7x\)
Show Answer
We can use the logarithm formula in order to determine the derivative:
\(\cfrac{d}{dx}\cdot [a \cdot \text{log}_b(x)] = \cfrac{a}{x\cdot \text{ln}(b)}\)
In this equation, we can identify that \(a = -3\) and \(b = 7\). As a result, we can express the derivative as:
\(y' = \cfrac{\textcolor{red}{-3}}{\textcolor{blue}{x\cdot \text{ln}(7)}}\)
Therefore, we can determine that \(y' = \cfrac{-3}{x\cdot \text{ln}(7)}\).
\(y = \cfrac{\text{ln}x}{x^3}\)
Show Answer
We can use Quotient Rule to determine the derivative:
First, we can determine the respective derivatives for \(f(x)\) and \(g(x)\). We can use Natural Logarithm Rule to determine \(f'(x)\):
\(\cfrac{d}{dx}\cdot \text{ln}(x) = \cfrac{1}{x}\)
\(\textcolor{red}{f'(x) = \cfrac{1}{x}}\)
\(g'(x) = 3(x^2)\)
\(\textcolor{blue}{g'(x) = 3x^2}\)
Next, we can implement Quotient Rule to determine the derivative:
\(h'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(y' = \cfrac{\left[\textcolor{blue}{x^3}\cdot \textcolor{red}{\cfrac{1}{x}}\right] - [\textcolor{red}{\ln(x)}\cdot \textcolor{blue}{3x^2}]}{(\textcolor{blue}{x^3})^2}\)
\(y' = \cfrac{x^2 - [3x^2 \cdot \text{ln}(x)]}{x^6}\)
Then, we can take a common factor (\(x^2\)) out of the numerator and simplify to get our final derivative:
\(y' = \cfrac{\cancel{x^2}[1 - 3\cdot \text{ln}(x)]}{x^{\cancel{6}4}}\)
\(y' = \cfrac{1 - 3\cdot \text{ln}(x)}{x^4}\)
Therefore, we can determine that \(y' = \cfrac{1 - 3\cdot \text{ln}(x)}{x^4}\).