Today I woke up and forgot to take the chicken I'm having for dinner out of the freezer. What should I do? Common advice is to sink the package of chicken into water so it thaws faster. Of course I'm not having chicken with water for dinner! I'm just using the water as a between-step so my dinner is not frozen!
Now think of the chicken as a function, my dinner would be its derivative. What can be the water in my above predicament? Anything that makes differentiation easier but it's not present in the end result! In this article you will explore how to use logarithms to find derivatives.
Logarithmic Differentiation Rule
Logarithms have unique properties like the Product Property of Logarithms and the Power Property of Logarithms, to name a few. These properties can be used when finding the derivative of more complex functions. This is done by using Logarithmic Differentiation, which more of a method than a rule.
Logarithmic Differentiation is a method that finds the derivative of the logarithm of the function rather than the original function.
But why do you need to use logarithmic differentiation? To take advantage of the properties of logarithms of course!
The properties of logarithms as a bridge to simpler operations.
It is time to take a look at the steps for doing Logarithmic Differentiation.
Logarithmic Differentiation Steps
The method of Logarithmic Differentiation can be summarized in the following steps:
Take the natural logarithm of the original function.
Use any relevant properties of logarithms, like the Power Property of Logarithms or the Product Property of Logarithms. The goal of this step is to simplify the function.
Use the Chain Rule and the differentiation rule of the natural logarithm to differentiate each expression.
Multiply the resulting expression by the original function. The result is the derivative of the original function.
The second step is where you can take advantage of the logarithmic differentiation. The properties of logarithms will help you simplify the operations required.
These steps are better understood with examples. Let's dig in!
Logarithmic Differentiation Examples
You can use Logarithmic Differentiation in a wide variety of situations. The properties of logarithms can help you simplify the process of finding the derivative of a function. These can be classified according to which property of logarithms is used to simplify the expressions.
Find the derivative of the function
\[ f(x)=x^8 e^x.\]
Answer:
Before starting please note that you can also use the Product Rule to find the derivative of this function. This example illustrates how to use logarithmic differentiation to obtain the same answer.
1. Take the natural logarithm of the original function.
Begin by taking the natural logarithm of the function, so
\[ \ln{f(x)}=\ln{\left( x^8 e^x \right)}.\]
2. Use any relevant properties of logarithms. In this case, the product property of logarithms and the power property of logarithms.
Since the right-hand side of the equation is the logarithm of a product, it can be written as the sum of logarithms, that is
\[ \ln{f(x)}= \ln{x^8} + \ln{e^x}.\]
Furthermore, you can use the power property of logarithms to write each exponent as a factor, obtaining
\[ \begin{align} \ln{f(x)} &= 8\ln{x} +x\ln{e} \\ &= 8\ln{x}+x, \end{align}\]
where you have also used the fact that \(\ln{e}=1.\)
3. Differentiate each expression.
Next, you need to differentiate both sides of the above expression with the help of the Chain Rule, the Power Rule, and the differentiation rule for the natural logarithm,
\[ \frac{\mathrm{d}}{\mathrm{d}x}\ln{x}=\frac{1}{x},\]
obtaining
\[ \begin{align} \left( \frac{1}{f(x)} \right) \left( f'(x) \right) &= \frac{8}{x}+1 \\ \frac{f'(x)}{f(x)} &= \frac{8}{x}+1. \end{align}\]
4. Multiply the resulting expression by the original function.
Finally, isolate the derivative by multiplying both sides of the above expression by the original function, \( f(x)=x^8 e^x,\) and simplify, that is
\[ \begin{align} f'(x) &= f(x)\left( \frac{8}{x}+1 \right) \\[0.5em] &= x^8e^x\left( \frac{8}{x}+1 \right) \\[0.5em] &= e^x\left( \frac{8x^8}{x} +x^8 \right) \\[0.5em] &= e^x(8x^7+x^8). \end{align} \]
Notice that this is exactly what you expected to get!
What about the quotient property of logarithms?
Find the derivative of
\[g(x)=\frac{\sqrt{x+1}}{x^2}.\]
Answer:
Rather than using the Quotient Rule (which sometimes is hard to remember) you can use Logarithmic Differentiation!
1. Take the natural logarithm of the original function.
This step is rather straightforward, doing so gives you
\[\ln{g(x)} = \ln{\left( \frac{\sqrt{x+1}}{x^2} \right)}.\]
2. Use any relevant properties of logarithms. In this case use the quotient property of logarithms and the power property of logarithms.
The logarithm of the quotient can be written as a difference of logarithms, that is
\[ \ln{g(x)} = \ln{\sqrt{x+1}}-\ln{x^2}. \]
Also, you can write the powers (remember that a square root is a power of \( ^1/_2 \) ) as factors using the power property of logarithms, so
\[ \ln{g(x)} = \frac{1}{2}\ln{\left(x+1 \right)}-2\ln{x}.\]
3. Differentiate each expression.
This time differentiating both sides of the above expression gives you
\[ \begin{align} \frac{g'(x)}{g(x)} &= \frac{1}{2}\cdot \frac{1}{x+1}-2\cdot\frac{1}{x} \\ &= \frac{1}{2}\cdot \frac{1}{x+1} -\frac{2}{x}, \end{align} \]
which can be simplified by adding the rational expressions
\[ \frac{g'(x)}{g(x)}= \frac{-3x-4}{2x(x+1)}. \]
4. Multiply the resulting expression by the original function.
Isolate the derivative by multiplying both sides of the above expression by \( g(x) \) and simplify, that is
\[ \begin{align} g'(x) &= \left( g(x)\right) \left(\frac{-3x-4}{2x(x+1)}\right) \\[0.5em] &= \left( \frac{\sqrt{x+1}}{x^2} \right) \left( \frac{-3x-4}{2x(x+1)}\right) \\[0.5em] &= \frac{-3x-4}{2x^3\sqrt{x+1}}. \end{align}\]
Logarithmic Differentiation can be used to find the derivative of a very peculiar function.
Find the derivative of
\[h(x)=x^x.\]
Answer:
Here you have \(x\) raised to the power of \(x.\) You identify an exponential function when the variable is the power and not the base, and the Power Rule only applies if the variable is not in the exponent. In this case the variable is both the base and the power! What to do? Logarithmic differentiation of course!
1. Take the natural logarithm of the original function.
As usual, begin by taking the natural logarithm of the function, that is
\[ \ln{h(x)} = \ln{x^x}.\]
2. Use any relevant properties of logarithms. In this case use the power property of logarithms.
You can now rewrite the power as a factor using the power property of logarithms, giving you
\[ \ln{h(x)} = (x)(\ln{x}). \]
3. Differentiate each expression.
The right-hand side of the above expression is a product of functions, hence it can be differentiated with the product rule, so
\[\begin{align} \frac{h'(x)}{h(x)} &= \left(\frac{\mathrm{d}}{\mathrm{d}x} x \right)\ln{x} + x\left(\frac{\mathrm{d}}{\mathrm{d}x}\ln{x}\right) \\ &= (1)(\ln{x}) + x\left( \frac{1}{x} \right) \\ &= \ln{x}+1. \end{align}\]
4. Multiply the resulting expression by the original function.
Finally, isolate the derivative by multiplying both sides of the above expression by \( h(x) \)
\[ \begin{align} h'(x) &= \left( h(x) \right) \left( \ln{x}+1 \right) \\ &= \left(x^x \right) \left( \ln{x}+1 \right). \end{align} \]
As you can see, logarithmic differentiation is very useful for avoiding working with larger expressions or finding the derivatives of functions that cannot be worked using standard differentiation techniques.
Using Logarithmic Differentiation to obtain Formulas
Logarithmic differentiation can also be used to prove some differentiation rules, like the Product Rule and the Quotient Rule. Let's dive into their proof using Logarithmic Differentiation!
You can prove the Product Rule using Logarithmic Differentiation. Consider the function
\[f(x)=g(x)h(x).\]
As usual, begin by taking the natural logarithm of both sides of the function rule
\[ \ln{f(x)} = \ln{\left( g(x)h(x) \right)}, \]
which you can rewrite using the power property of logarithms in the right-hand side, so
\[ \ln{f(x)} = \ln{g(x)} + \ln{h(x)}. \]
Now you can differentiate both sides of the equation using the Chain Rule, that is
\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g(x)}+\frac{h'(x)}{h(x)}. \]
Finally, multiply the equation by \( f(x) \)
\[ \begin{align} f'(x) &= f(x)\left( \frac{g'(x)}{g(x)}+\frac{h'(x)}{h(x)} \right) \\ &= g(x)h(x)\frac{g'(x)}{g(x)} + g(x)h(x)\frac{h'(x)}{h(x)} \\ &= h(x)g'(x)+g(x)h'(x). \end{align} \]
The above expression is the Product Rule we all know! You can try proving the Quotient Rule using a procedure similar as above.
For which calculations you should use Logarithmic Differentiation?
Since the goal of using Logarithmic Differentiation is to simplify the process of finding the derivative of a function you should only use it when the derivative becomes easier to find. It can also be used when the derivative of a function cannot be found with standard differentiation techniques, like with \(f(x)=x^x.\)
Whenever you can use the Product Rule or the Quotient Rule, you can also use Logarithmic Differentiation. While the Product Rule might be easier to work around, sometimes you might forget which is the negative term of the Quotient Rule.
There is one common mistake when using the Quotient Rule and that is getting the signs mistaken.
\[\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)} \neq \frac{f(x)g'(x)-g(x)f'(x)}{\left( g(x) \right)^2}\]
You can prevent this mistake by using Logarithmic Differentiation since it is easier to recall that the negative term is the one in the denominator.
\[ \ln{\left(\frac{f(x)}{g(x)} \right)} = \ln{f(x)}-\ln{g(x)}.\]
From here, you can continue the process of Logarithmic Differentiation to find the derivative of the function.
\[\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)} = \frac{g(x)f'(x)-f(x)g'(x)}{\left( g(x) \right)^2} \]
Logarithmic Differentiation - Key takeaways
- Logarithmic Differentiation is a method used to find derivatives using the properties of logarithms.
- The steps followed for Logarithmic Differentiation are the following:
- Take the natural logarithm of the original function.
- Use any relevant properties of logarithms to simplify the function.
- Use the Chain Rule and the differentiation rule of the natural logarithm to differentiate the expression.
- Multiply the resulting expression by the original function.
- The following properties of logarithms can be used to your favor when simplifying expressions:
- Product Property of Logarithms.
- Quotient Property of Logarithms.
- Power Property of Logarithms.
- Logarithmic Differentiation should be used when the derivative becomes easier to find. There is no point in using Logarithmic Differentiation to find the derivative of \(f(x)=x^n.\)
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