Logarithmic Differentiation #
Let’s say you have an expression like \( x^x \).
You can’t use the regular power rule here because the exponent isn’t a constant. So what do we do?
We use logarithmic differentiation.
First Example: \( x^x \) #
Let:
\[ y = x^x \]Take the natural log (ln) of both sides. You can technically use any base, but base \( e \) (i.e., \( \ln \)) is the most convenient.
\[ \ln y = \ln(x^x) \]Using log rules to move the exponent down:
\[ \ln y = x \ln x \]Now, differentiate both sides. Remember the rule:
\[ \frac{d}{dx}[\ln(f(x))] = \frac{f'(x)}{f(x)} \]So:
\[ \frac{d}{dx}[\ln y] = \frac{dy}{dx} \cdot \frac{1}{y} = \frac{y'}{y} \]And for the right-hand side \( x \ln x \), we use the product rule:
\[ \frac{d}{dx}[x \ln x] = x \cdot \frac{1}{x} + \ln x = 1 + \ln x \]Putting it all together:
\[ \frac{y'}{y} = 1 + \ln x \]Now multiply both sides by \( y \):
\[ y' = y(1 + \ln x) \]Since \( y = x^x \), we substitute:
\[ \boxed{y' = x^x(1 + \ln x)} \]Harder Example: \( (\sin x)^x \) #
Let:
\[ y = (\sin x)^x \]Again, take the natural log of both sides:
\[ \ln y = \ln\left((\sin x)^x\right) \]Use the same trick: bring the exponent down:
\[ \ln y = x \ln(\sin x) \]Now differentiate both sides:
- Left-hand side:
- Right-hand side (product rule):
Use the chain rule on \( \ln(\sin x) \):
\[ \frac{d}{dx}[\ln(\sin x)] = \frac{\cos x}{\sin x} = \cot x \]So now the derivative becomes:
\[ \frac{y'}{y} = x \cot x + \ln(\sin x) \]Multiply both sides by \( y \):
\[ y' = y(x \cot x + \ln(\sin x)) \]And substitute \( y = (\sin x)^x \):
\[ \boxed{y' = (\sin x)^x (x \cot x + \ln(\sin x))} \]Summary #
Logarithmic differentiation is your go-to tool when:
- The exponent is not constant
- You have variable bases and exponents
- You need to simplify messy products involving exponents and logs
Credits #
All explanations and problems written by Adrian Hernandez Vega, unless otherwise noted.
Prefer video? Watch the full explanation here: