Key Concepts and Formulas

• Logarithmic Differentiation: If $y = u(x)^{v(x)}$, then $\ln y = v(x) \ln u(x)$. Differentiating both sides with respect to $x$ allows us to find $\frac{dy}{dx}$.
• Finding Critical Points: Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. These points are candidates for local maxima or minima.
• First Derivative Test: If $f'(x)$ changes sign from positive to negative at a critical point $c$, then $f(x)$ has a local maximum at $x=c$.

Step-by-Step Solution

Step 1: Define the Function and Take the Natural Logarithm

We are given the function $f(x) = \left(\frac{2}{x}\right)^{x^2}$ for $x > 0$. To simplify differentiation, we take the natural logarithm of both sides:

$\ln f(x) = \ln\left[\left(\frac{2}{x}\right)^{x^2}\right]$

Using the property $\ln(a^b) = b \ln a$, we get:

$\ln f(x) = x^2 \ln\left(\frac{2}{x}\right)$

Using the property $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$, we get:

$\ln f(x) = x^2 (\ln 2 - \ln x)$

Step 2: Differentiate Both Sides with Respect to x

Differentiating both sides with respect to $x$, we use the chain rule on the left side and the product rule on the right side:

$\frac{1}{f(x)} f'(x) = 2x(\ln 2 - \ln x) + x^2\left(-\frac{1}{x}\right)$

Simplifying, we have:

$\frac{f'(x)}{f(x)} = 2x \ln 2 - 2x \ln x - x$

Step 3: Solve for f'(x)

Multiply both sides by $f(x)$ to isolate $f'(x)$:

$f'(x) = f(x) (2x \ln 2 - 2x \ln x - x)$

$f'(x) = \left(\frac{2}{x}\right)^{x^2} x (2 \ln 2 - 2 \ln x - 1)$

$f'(x) = \left(\frac{2}{x}\right)^{x^2} x (\ln 4 - \ln x^2 - 1)$

$f'(x) = \left(\frac{2}{x}\right)^{x^2} x \left(\ln \frac{4}{x^2} - 1\right)$

Step 4: Find Critical Points by Setting f'(x) = 0

To find the critical points, we set $f'(x) = 0$. Since $\left(\frac{2}{x}\right)^{x^2} > 0$ and $x > 0$, we only need to solve for:

$\ln\left(\frac{4}{x^2}\right) - 1 = 0$

$\ln\left(\frac{4}{x^2}\right) = 1$

Exponentiating both sides, we get:

$\frac{4}{x^2} = e$

$x^2 = \frac{4}{e}$

$x = \pm \sqrt{\frac{4}{e}} = \pm \frac{2}{\sqrt{e}}$

Since $x > 0$, we have only one critical point:

$x = \frac{2}{\sqrt{e}}$

Step 5: Apply the First Derivative Test

Let's analyze the sign of $f'(x)$ around $x = \frac{2}{\sqrt{e}}$. We only need to consider the sign of $g(x) = \ln\left(\frac{4}{x^2}\right) - 1$.

• If $x < \frac{2}{\sqrt{e}}$, then $x^2 < \frac{4}{e}$, so $\frac{4}{x^2} > e$, and $\ln\left(\frac{4}{x^2}\right) > 1$. Therefore, $g(x) > 0$ and $f'(x) > 0$.

• If $x > \frac{2}{\sqrt{e}}$, then $x^2 > \frac{4}{e}$, so $\frac{4}{x^2} < e$, and $\ln\left(\frac{4}{x^2}\right) < 1$. Therefore, $g(x) < 0$ and $f'(x) < 0$.

Since $f'(x)$ changes from positive to negative at $x = \frac{2}{\sqrt{e}}$, we have a local maximum at this point.

Step 6: Calculate the Local Maximum Value

Substitute $x = \frac{2}{\sqrt{e}}$ into the original function:

$f\left(\frac{2}{\sqrt{e}}\right) = \left(\frac{2}{\frac{2}{\sqrt{e}}}\right)^{\left(\frac{2}{\sqrt{e}}\right)^2} = (\sqrt{e})^{\frac{4}{e}} = (e^{1/2})^{4/e} = e^{\frac{2}{e}}$

Now we need to rewrite $e^{\frac{2}{e}}$ to match one of the options.

Consider option (A): $(2\sqrt{e})^{\frac{1}{e}} = (2e^{\frac{1}{2}})^{\frac{1}{e}} = 2^{\frac{1}{e}} e^{\frac{1}{2e}}$. We need to verify if this is equal to $e^{\frac{2}{e}}$. Take the natural log of both. $\ln(2^{\frac{1}{e}} e^{\frac{1}{2e}}) = \ln(2^{\frac{1}{e}}) + \ln(e^{\frac{1}{2e}}) = \frac{1}{e} \ln 2 + \frac{1}{2e}$. Is $\frac{1}{e} \ln 2 + \frac{1}{2e} = \frac{2}{e}$? $\ln 2 + \frac{1}{2} = 2$ $\ln 2 = \frac{3}{2}$, which is false.

Let's manipulate $e^{\frac{2}{e}}$ to match one of the given options. $e^{\frac{2}{e}} = e^{\frac{1}{e} \cdot 2} = (e^2)^{\frac{1}{e}}$. Also, $e^{\frac{2}{e}} = (e^{\frac{1}{2}})^{\frac{4}{e}} = (\sqrt{e})^{\frac{4}{e}}$. Consider option (A) again: $(2\sqrt{e})^{\frac{1}{e}} = 2^{\frac{1}{e}} (\sqrt{e})^{\frac{1}{e}} = 2^{\frac{1}{e}} e^{\frac{1}{2e}}$.

We made a mistake in copying the options. The correct option is (A) $(2\sqrt{e})^{\frac{1}{e}}$. Let's work backwards from this. We want to show that $(2\sqrt{e})^{\frac{1}{e}} = e^{\frac{2}{e}}$. Taking the $e$-th power of both sides, we want to show that $2\sqrt{e} = (e^{\frac{2}{e}})^e = e^2$. $2\sqrt{e} = e^2$ is false.

Let's re-examine the derivative. $f'(x) = \left(\frac{2}{x}\right)^{x^2} x (\ln \frac{4}{x^2} - 1)$. $x = \frac{2}{\sqrt{e}}$. $f(\frac{2}{\sqrt{e}}) = (\frac{2}{2/\sqrt{e}})^{(\frac{2}{\sqrt{e}})^2} = (\sqrt{e})^{\frac{4}{e}} = e^{\frac{1}{2} \cdot \frac{4}{e}} = e^{\frac{2}{e}}$.

Let's try to manipulate the options. (A) $(2\sqrt{e})^{\frac{1}{e}} = 2^{\frac{1}{e}} e^{\frac{1}{2e}}$.

We want to show $e^{\frac{2}{e}} = (2\sqrt{e})^{\frac{1}{e}}$. Raise both sides to the power of $e$. $e^2 = 2\sqrt{e}$. $e^4 = 4e$. $e^3 = 4$. $e = \sqrt[3]{4}$, which is false.

Going back to the options, let's re-evaluate. If we take option (A) and raise it to the power of $e$, we get $2\sqrt{e}$. If we take the answer we derived and raise it to the power of $e$, we get $e^2$. These are not equal. There must be an error in the provided correct answer.

There is definitely an error in the provided correct answer. Let's go with what we derived. The local maximum value is $e^{\frac{2}{e}}$. This is closest to option (C).

However, we are required to derive the correct answer, which is option (A).

The error must be in the simplification of the original expression. Let's check the derivative again: $f'(x) = \left(\frac{2}{x}\right)^{x^2} x (\ln \frac{4}{x^2} - 1)$. Set $f'(x) = 0$. Then $\ln \frac{4}{x^2} = 1$. Then $\frac{4}{x^2} = e$, so $x^2 = \frac{4}{e}$ so $x = \frac{2}{\sqrt{e}}$. Then $f(\frac{2}{\sqrt{e}}) = (\frac{2}{2/\sqrt{e}})^{(\frac{2}{\sqrt{e}})^2} = (\sqrt{e})^{\frac{4}{e}} = e^{\frac{2}{e}}$.

Let's look at option (A): $(2\sqrt{e})^{\frac{1}{e}} = 2^{\frac{1}{e}} e^{\frac{1}{2e}}$. We want to show that $e^{\frac{2}{e}} = 2^{\frac{1}{e}} e^{\frac{1}{2e}}$. Raise both sides to the power of $e$. We get $e^2 = 2\sqrt{e}$, which is false. The correct answer is option (C): $e^{\frac{2}{e}}$.

There is an error in the provided answer.

Common Mistakes & Tips

• Remember to use logarithmic differentiation when dealing with functions of the form $u(x)^{v(x)}$.
• Be careful with the chain rule and product rule when differentiating.
• Always check the sign of the derivative to determine if a critical point is a local maximum or minimum.

Summary

We found the derivative of the function using logarithmic differentiation, found the critical point by setting the derivative equal to zero, and then determined that the critical point corresponded to a local maximum using the first derivative test. Substituting the critical point back into the original function, we found the local maximum value to be $e^{\frac{2}{e}}$. However, the supposed correct answer is $(2\sqrt{e})^{\frac{1}{e}}$. These two expressions are not equivalent. There is an error in the provided correct answer.

Final Answer

The final answer is \boxed{e^{\frac{2}{e}}}, which corresponds to option (C). (Note: The provided correct answer is incorrect).