Animated Solution for Mathematics - Differentiation: If $f(x)={\log }_x(\ln x)$, then $f^{\prime }(x)$ at $x=e$ is $\dots \dots \dots$

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that often makes students pause in their tracks: $f(x)={\log }_x(\ln x)$.

At first glance, it looks intimidating. We are used to logarithms with constant bases, like ${\log }_{10}x$ or $\ln x$. But here, the base itself is a variable, $x$.

This is the moment where many students feel the urge to panic, but I want you to take a deep breath. In calculus, when a function looks unfamiliar, our goal is to translate it into a language we already speak.

Standard differentiation rules, like the power rule or the basic logarithmic derivative, do not apply directly when the base is a function of $x$. We need a translator. That translator is the Change of Base Formula: ${\log }_{a}b=\frac{\ln b}{\ln a}$.

By applying this, we transform our function into something much more manageable:

Suddenly, the mystery vanishes. We are no longer dealing with a strange logarithmic base; we are dealing with a simple quotient of two functions. This is the beauty of mathematics—with the right identity, even the most complex-looking expression can be tamed.

Now that we have $f(x)=\frac{\ln (\ln x)}{\ln x}$, we recognize the structure. It is a ratio, $u/v$. To find the derivative, we must summon the Quotient Rule:

Let's define our components: - $u=\ln (\ln x)$ - $v=\ln x$

To proceed, we need the derivatives $u^{\prime }$ and $v^{\prime }$. The derivative of the denominator, $v^{\prime }=\frac{d}{dx}(\ln x)$, is a classic: $v^{\prime }=\frac{1}{x}$.

The numerator, $u^{\prime }$, requires the Chain Rule. We differentiate the outer natural log, which gives $\frac{1}{\ln x}$, and multiply it by the derivative of the inner $\ln x$, which is $\frac{1}{x}$. Thus, $u^{\prime }=\frac{1}{x\ln x}$.

Now, we assemble our derivative using the Quotient Rule framework:

I know, it looks like a mess of symbols. But look closely at the first term in the numerator: $(\ln x)\cdot \left(\frac{1}{x\ln x}\right)$. The $\ln x$ terms cancel out perfectly, leaving us with just $\frac{1}{x}$.

Now our expression is:

We can factor out $\frac{1}{x}$ from the numerator, which gives us:

We have arrived at the final stage. We need to evaluate this derivative at $x=e$. This is where the magic happens. Substituting $x=e$ into our expression:

Recall that $\ln e=1$. So, the expression becomes:

Since $\ln (1)=0$, the numerator simplifies to $1-0=1$. The denominator is simply $e\cdot 1=e$. Therefore, the final result is:

And there it is. A complex, variable-based logarithmic function yields a clean, elegant result. This is the reward for your patience and your methodical approach. You didn't just solve a problem; you navigated a path through the logic of calculus. Keep this confidence with you as you face the next challenge!