Can I use the power rule for $x^x$?

No. The power rule $nx^{n-1}$ only applies when the exponent is a real constant. Variable exponents require logarithmic differentiation or rewriting the base as $e$.

What if the base function $f(x)$ is negative?

Logarithmic differentiation requires $f(x) > 0$ because the domain of the real-valued natural logarithm is strictly positive. For intervals where $f(x) < 0$, a real-valued variable exponent function is generally undefined.

Can this method be used for complex products and quotients?

Yes. Logarithmic differentiation is highly efficient for simplifying the differentiation of large fractions or multiple multiplied terms, as it converts products to sums and quotients to differences.

How to differentiate a function with a variable exponent using logarithmic differentiation

Q: Can this method be used for complex products and quotients?

Yes. Logarithmic differentiation is highly efficient for simplifying the differentiation of large fractions or multiple multiplied terms, as it converts products to sums and quotients to differences.

Logarithmic differentiation transforms functions of the form $y = [f(x)]^{g(x)}$ into manageable products by applying the natural logarithm to both sides. This method applies whenever the base and the exponent are both non-constant functions of the independent variable, where standard power and exponential rules fail.

The setup

Identify a function of the form $y = f(x)^{g(x)}$ where $f(x) > 0$ . Begin by applying the natural logarithm to both sides of the equation, yielding $\ln(y) = \ln(f(x)^{g(x)})$ . Use the logarithm power rule to rewrite the right side as a product: $\ln(y) = g(x) \ln(f(x))$ .

The steps

Apply $\ln$ to both sides of $y = f(x)^{g(x)}$ . 2. Rewrite the right side as $g(x)\ln(f(x))$ . 3. Differentiate both sides implicitly with respect to $x$ . The left side becomes $\frac{1}{y} \frac{dy}{dx}$ . The right side requires the product rule and chain rule. 4. Multiply both sides of the equation by $y$ to isolate $\frac{dy}{dx}$ . 5. Substitute the original expression for $y$ back into the right side to express the final derivative strictly in terms of $x$ .

Checking the result

Verify that the final derivative contains the original function $f(x)^{g(x)}$ as a leading factor. You can cross-check the result by rewriting the original function using base $e$ , such that $y = e^{g(x)\ln(f(x))}$ , and differentiating using the standard exponential chain rule. Both methods must yield identical expressions.

Common errors

A frequent error is forgetting to multiply the differentiated right side by $y$ in the final step. Another common mistake is applying the standard power rule to $f(x)^{g(x)}$ , incorrectly yielding $g(x)f(x)^{g(x)-1}$ , which is mathematically invalid for variable exponents. Finally, students often misapply the product rule when differentiating $g(x)\ln(f(x))$ .

Worked example

Find the derivative of $y = x^{\sin(x)}$ with respect to $x$ .

Let $y = x^{\sin(x)}$ . Apply the natural logarithm to both sides: $\ln(y) = \ln(x^{\sin(x)})$ . Use log properties to bring down the exponent: $\ln(y) = \sin(x) \ln(x)$ . Differentiate both sides with respect to $x$ . The left side is $\frac{1}{y} \frac{dy}{dx}$ . The right side uses the product rule: $\frac{d}{dx}[\sin(x)] \ln(x) + \sin(x) \frac{d}{dx}[\ln(x)] = \cos(x)\ln(x) + \sin(x)\frac{1}{x}$ . Equate the two sides: $\frac{1}{y} \frac{dy}{dx} = \cos(x)\ln(x) + \frac{\sin(x)}{x}$ . Multiply by $y$ : $\frac{dy}{dx} = y \left[ \cos(x)\ln(x) + \frac{\sin(x)}{x} \right]$ . Substitute $y = x^{\sin(x)}$ back into the equation: $\frac{dy}{dx} = x^{\sin(x)} \left[ \cos(x)\ln(x) + \frac{\sin(x)}{x} \right]$ .

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References: Stewart Calculus, 8th Edition, Section 3.6 · OpenStax Calculus Volume 1, Chapter 3.9