Animated Solution for Mathematics - Differentiation: Use the function $f(x)=x^{1/x},x>0$, to determine the bigger of the two numbers $e^{\pi }$ and ${\pi }^e$.

To compare $e^{\pi }$ and ${\pi }^e$, we need to bring them into a common framework. If we take the $e\pi$-th root of both numbers, we transform the problem.

The expression $(e^{\pi }{)}^{\frac{1}{e\pi }}$ simplifies beautifully to $e^{\frac{1}{e}}$, and $({\pi }^e{)}^{\frac{1}{e\pi }}$ becomes ${\pi }^{\frac{1}{\pi }}$.

Now, the problem is no longer about two arbitrary titans; it is about comparing two values of a single, beautiful function: $f(x)=x^{\frac{1}{x}}$.

To understand the behavior of $f(x)=x^{\frac{1}{x}}$, we must look at its rate of change. We define $y=x^{\frac{1}{x}}$.

Because the variable $x$ is trapped in both the base and the exponent, we use logarithmic differentiation. Taking the natural logarithm on both sides, we get $\ln y=\frac{1}{x}\ln x$.

Now, we differentiate with respect to $x$. Using the chain rule on the left and the quotient rule on the right, we find:

Simplifying this, we arrive at the derivative:

This derivative is the key to the entire kingdom.

To find the peaks and valleys of our function, we set $f^{\prime }(x)=0$. Since $x^{\frac{1}{x}}$ and $x^2$ are always positive for $x>0$, the only way for the derivative to be zero is if $1-\ln x=0$.

This leads us directly to $\ln x=1$, or $x=e$. This is the critical point. It is the peak of the function $f(x)$.

Now, let us look at what happens when $x>e$. For any $x>e$, we know that $\ln x>1$.

Consequently, the term $(1-\ln x)$ becomes negative. This means that for all $x>e$, the derivative $f^{\prime }(x)$ is negative.

A negative derivative tells us that the function is strictly decreasing. We know that $e\approx 2.718$ and $\pi \approx 3.141$. Since $\pi >e$, and the function is strictly decreasing in this interval, it must be that $f(\pi )<f(e)$.

We have established that $f(\pi )<f(e)$, which means ${\pi }^{\frac{1}{\pi }}<e^{\frac{1}{e}}$. To return to our original numbers, we raise both sides to the power of $e\pi$.

The inequality holds, and we find that $e^{\pi }>{\pi }^e$.

The mystery is solved. Through the power of calculus, we have tamed the titans. Remember, in mathematics, the most complex problems often yield to the most elegant insights.