Logarithmic Differentiation

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!

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

\[ \begin{align} g'(x) &= \left( g(x)\right) \left(\frac{-3x-4}{2x(x+1)}\right) \\[0.5em] &= \left( \frac{\sqrt{x+1}}{x^2} \right) \left( \frac{-3x-4}{2x(x+1)}\right) \\[0.5em] &= \frac{-3x-4}{2x^3\sqrt{x+1}}. \end{align}\]

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 \( h(x) \)

\[ \begin{align} h'(x) &= \left( h(x) \right) \left( \ln{x}+1 \right) \\ &= \left(x^x \right) \left( \ln{x}+1 \right). \end{align} \]

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.

\[\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)} = \frac{g(x)f'(x)-f(x)g'(x)}{\left( g(x) \right)^2} \]