Q.Find $\frac{dy}{dx}$ in the following: $x^x + x^a + a^x + a^a$, for some fixed $a > 0$ and $x > 0$

The key idea is Implicit Differentiation combined with logarithmic differentiation for variable-exponent terms.

Step 1: Let $y = x^x + x^a + a^x + a^a$. Since $a$ is a fixed constant, $\frac{d}{dx}(a^a) = 0$.

Step 2: For $x^x$, take logs: $\log(x^x) = x \log x$. Differentiate: $\frac{d}{dx}(x^x) = x^x (1 \cdot \log x + x \cdot \frac{1}{x}) = x^x (\log x + 1)$.

Step 3: For $x^a$, use the power rule: $\frac{d}{dx}(x^a) = a x^{a-1}$. …

We are given $y = x^x + x^a + a^x + a^a$, where $a > 0$ is a fixed constant and $x > 0$. The goal is to find $\frac{dy}{dx}$.

The key idea is that each term is a different type of function, so each requires its own differentiation technique. The constant term $a^a$ vanishes. The term $x^a$ is a power function (variable base, constant exponent), $a^x$ is an exponential function (constant base, variable exponent), and $x^x$ is a "variable base, variable exponent" — which needs logarithmic differentiation.

Let’s go term by term.

1. The constant term $a^a$

   Since $a$ is fixed, $a^a$ is just a number. Its derivative is $0$.

2. The power term $x^a$

   Here the exponent $a$ is constant. This is a standard power rule:

3. The exponential term $a^x$ Here the base $a$ is constant. The derivative of $a^x$ is $a^x \log a$.

4. The tricky term $x^x$ Both base and exponent depend on $x$. The standard trick: take the natural logarithm of both sides, differentiate implicitly, then solve for the derivative. Let $u = x^x$. Then $\log u = x \log x$. Differentiate both sides with respect to $x$:

Multiply through by $u = x^x$: …

Method: Differentiating a Sum with Mixed Power/Exponential/Variable-Power Terms

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 $y$

Step 4: Substitute the original expression for $y$ back in …

Common Mistakes

Mistake 1: Applying logarithmic differentiation to every term instead of only $x^x$.

Why it's wrong: $x^a$ and $a^x$ already have a constant in one position (exponent or base respectively) and are handled faster and more reliably by the plain power rule / exponential rule — over-applying log differentiation wastes time and introduces unnecessary risk of error. Correct approach: classify each term's structure first, then pick the matching rule.

Mistake 2: Guessing the derivative of $x^x$ as either $x\cdot x^{x-1}=x^x$ (treating it like a power) or $x^x\log x$ (treating it like an exponential). …