Key Concepts and Formulas

• First Derivative Test: A function $f(x)$ is strictly increasing on an interval $(a, b)$ if $f'(x) > 0$ for all $x$ in $(a, b)$.
• Logarithmic Differentiation: If $f(x) = u(x)^{v(x)}$, then $\ln(f(x)) = v(x) \ln(u(x))$. Differentiating both sides implicitly can simplify finding $f'(x)$. Alternatively, use $a^b = e^{b \ln a}$.
• Derivative of $e^{g(x)}$: $\frac{d}{dx}(e^{g(x)}) = e^{g(x)} \cdot g'(x)$

Step-by-Step Solution

Step 1: Rewrite the function using exponential and logarithmic properties.

We are given the function $f(x) = x^x$ for $x > 0$. To make differentiation easier, we rewrite the function using the identity $a^b = e^{b \ln a}$. Why? This transformation converts the exponentiation into a product involving the natural logarithm, which can be differentiated more easily using the chain and product rules. $f(x) = x^x = e^{x \ln x}$

Step 2: Differentiate the rewritten function.

We now need to find $f'(x)$. We will use the chain rule, noting that $f(x) = e^{g(x)}$ where $g(x) = x \ln x$. Why? The derivative $f'(x)$ will tell us where the function is increasing or decreasing. $f'(x) = \frac{d}{dx} \left(e^{x \ln x}\right) = e^{x \ln x} \cdot \frac{d}{dx} (x \ln x)$

Now, we need to find the derivative of $x \ln x$ using the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = x$ and $v = \ln x$. Then $u' = 1$ and $v' = \frac{1}{x}$.

$\frac{d}{dx}(x \ln x) = (1)(\ln x) + (x)\left(\frac{1}{x}\right) = \ln x + 1$

Substituting this back into the expression for $f'(x)$:

$f'(x) = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1)$

Step 3: Determine the condition for the function to be strictly increasing.

For $f(x)$ to be strictly increasing, we need $f'(x) > 0$. Why? The first derivative test states that a function is strictly increasing where its derivative is positive.

So, we need to find the values of $x$ for which $f'(x) = x^x (\ln x + 1) > 0$. Since $x^x > 0$ for all $x > 0$, we only need to consider the sign of $(\ln x + 1)$.

Therefore, we require: $\ln x + 1 > 0$

Step 4: Solve the inequality for $x$.

To find the interval where $f'(x) > 0$, we solve the inequality $\ln x + 1 > 0$. Why? This gives us the exact range of x values where the derivative is positive, hence the function is strictly increasing.

$\ln x > -1$

Exponentiating both sides with base $e$ (since $e^x$ is an increasing function, the inequality sign remains the same):

$e^{\ln x} > e^{-1}$ $x > \frac{1}{e}$

Thus, the function $f(x) = x^x$ is strictly increasing for $x > \frac{1}{e}$. In interval notation, this is $\left(\frac{1}{e}, \infty\right)$. Since we want the interval in which the function is strictly increasing, we should include the endpoint where $f'(x) = 0$. Since $f'(x) = 0$ at $x = \frac{1}{e}$, we include $\frac{1}{e}$ in the interval.

Therefore, the interval is $\left[\frac{1}{e}, \infty\right)$.

Common Mistakes & Tips

• Incorrect Differentiation: A common mistake is to try differentiating $x^x$ directly using the power rule, which is incorrect. Always use logarithmic differentiation or rewrite as $e^{x \ln x}$.
• Forgetting the Domain: Remember that the domain of $f(x) = x^x$ is $x > 0$, which is crucial for the problem.
• Sign Analysis: Pay close attention to the signs of the terms in $f'(x)$ when determining where it is positive or negative.

Summary

To find the interval where $f(x) = x^x$ is strictly increasing, we first rewrote the function as $f(x) = e^{x \ln x}$. Then, we found the derivative $f'(x) = x^x (\ln x + 1)$. Since $x^x > 0$ for $x > 0$, we only needed to solve the inequality $\ln x + 1 > 0$, which gave us $x > \frac{1}{e}$. Therefore, the function is strictly increasing on the interval $\left[\frac{1}{e}, \infty\right)$.

Final Answer The final answer is $\boxed{\left[\frac{1}{e}, \infty\right)}$, which corresponds to option (D).