Question

Differentiate the function $x^x$ with respect to the function $x \log x$.

Hint:

Notice during calculation that $\log u = v$. Differentiating both sides of $\log u = v$ with respect to $v$ gives $\frac{1}{u}\frac{du}{dv} = 1 \implies \frac{du}{dv} = u = x^x$. This alternative method is much faster!

Answer 1

Concept: When differentiating a function $u$ with respect to another function $v$, we utilize a variation of the chain rule. Instead of finding $\frac{du}{dx}$ directly, we compute the individual derivatives of both functions with respect to their shared underlying independent variable $x$, and then divide them: \[ \frac{du}{dv} = \frac{\left(\frac{du}{dx}\right)}{\left(\frac{dv}{dx}\right)} \] Since $u = x^x$ features a variable in both the base and the exponent, we must apply logarithmic differentiation to find its derivative.

Step 1: Differentiating $u = x^x$ with respect to $x$ using logarithmic differentiation.
Let: \[ u = x^x \] Taking the natural logarithm ($\log_e$ or $\ln$) on both sides: \[ \log u = \log\left(x^x\right) \] Using the logarithm power rule property $\log(a^b) = b \log a$: \[ \log u = x \log x \] Now, differentiate both sides with respect to $x$. Apply the chain rule on the left side and the product rule ($\frac{d}{dx}(f \cdot g) = f'g + fg'$) on the right side: \[ \frac{1}{u} \cdot \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) \] \[ \frac{1}{u} \cdot \frac{du}{dx} = 1 \cdot \log x + x \cdot \left(\frac{1}{x}\right) \] \[ \frac{1}{u} \cdot \frac{du}{dx} = \log x + 1 \] Multiply both sides by $u$ to isolate $\frac{du}{dx}$: \[ \frac{du}{dx} = u(1 + \log x) \] Substitute the original expression $u = x^x$ back into the equation: \[ \frac{du}{dx} = x^x(1 + \log x) \]

Step 2: Differentiating $v = x \log x$ with respect to $x$.
Let: \[ v = x \log x \] Differentiate $v$ with respect to $x$ using the standard product rule: \[ \frac{dv}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) \] \[ \frac{dv}{dx} = 1 \cdot \log x + x \cdot \left(\frac{1}{x}\right) \] \[ \frac{dv}{dx} = \log x + 1 \]

Step 3: Combining the derivatives using the function-by-function derivative formula.
Now, substitute the two individual derivatives from Step 1 and Step 2 into our core formula: \[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{x^x(1 + \log x)}{1 + \log x} \] Since $(1 + \log x)$ is common to both the numerator and the denominator, we can cancel them out: \[ \frac{du}{dv} = x^x \] Thus, the derivative of $x^x$ with respect to $x \log x$ is exactly $x^x$.