Q.Find $\frac{dy}{dx}$ in the following: $(\log x)^{\log x}, x > 1$

Concept: Implicit Differentiation — when the function is of the form $y = f(x)^{g(x)}$, take logs first to bring the exponent down, then differentiate implicitly.

Let $y = (\log x)^{\log x}$.

Take natural logarithms on both sides:

Differentiate both sides with respect to $x$ (implicitly on the left, product rule on the right):

Simplify the second term: $\log x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x}$. So: …

The core idea here is implicit differentiation — but why do we need it? The function $y = (\log x)^{\log x}$ has the variable $x$ in both the base and the exponent. Standard differentiation rules (like the power rule or the exponential rule) only handle one of these at a time. The power rule assumes a constant exponent; the exponential rule assumes a constant base. Here, both are changing with $x$, so we need a technique that untangles them.

Logarithmic differentiation is the perfect tool. By taking the natural logarithm of both sides, we convert the exponentiation into a product, which we can then differentiate using the chain rule and product rule. The logarithm "brings down" the exponent, making the relationship linear in terms of the logs.

Let’s work through it step by step.

1. Set up the equation.

   Let $y = (\log x)^{\log x}$, where $\log x$ denotes the natural logarithm (base $e$), and $x > 1$ ensures $\log x > 0$, so the expression is well-defined.

2. Take the natural logarithm of both sides.

   This is the key move:

Using the power property of logarithms, $\log(a^b) = b \log a$, we get:

Notice that $\log(\log x)$ is just the natural log of $\log x$ — don’t confuse it with $\log(\log x)$; they are the same here since $\log$ means natural log.

3. Differentiate both sides with respect to $x$. The left side: $\frac{d}{dx} (\log y) = \frac{1}{y} \cdot \frac{dy}{dx}$ (by the chain rule, since $y$ is a function of $x$). The right side: $\frac{d}{dx} \left[ (\log x) \cdot \log(\log x) \right]$ requires the product rule. Let $u = \log x$ and $v = \log(\log x)$. Then:

Compute $u'$: $u = \log x$, so $u' = \frac{1}{x}$.

Compute $v'$: $v = \log(\log x)$. Let $w = \log x$, then $v = \log w$, so $v' = \frac{1}{w} \cdot w' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.

Putting it together:

Simplify the second term: $(\log x) \cdot \frac{1}{x \log x} = \frac{1}{x}$. …

Method: Logarithmic Differentiation for $y=[f(x)]^{g(x)}$

When BOTH the base and the exponent of a power contain $x$, neither the power rule (needs constant exponent) nor the exponential rule (needs constant base) applies directly. Logarithmic differentiation converts the exponent into a product, which can then be handled with the ordinary rules.

Steps

Step 1: Take the natural log of both sides

using the power property of logs, $\log(a^b)=b\log a$.

Step 2: Differentiate both sides with respect to $x$

On the left, $\frac{d}{dx}\log y=\frac{1}{y}\frac{dy}{dx}$ (chain rule, since $y$ is a function of $x$). On the right, apply the product rule (since it's now $g(x)$ times $\log f(x)$), with the chain rule on $\log f(x)$ itself.

Step 3: Solve for $\dfrac{dy}{dx}$ by multiplying both sides by $y$ …

Common Mistakes

Mistake 1: Confusing $\log(\log x)$ with $(\log x)^2$ or attempting to "simplify" it further.

Why it's wrong: $\log(\log x)$ genuinely means the log of the log — it is not the same as squaring, and there is no further algebraic simplification available. Correct approach: treat $\log(\log x)$ as its own composite function and differentiate it via the chain rule, leaving it in that form.

Mistake 2: Forgetting to multiply the final expression by $y=(\log x)^{\log x}$ after solving for $\frac{1}{y}\frac{dy}{dx}$. …