Q.Find $\dfrac{dy}{dx}$, if $y^x + x^y + x^x = a^b$, where $a$ and $b$ are positive constants.

Concept: Implicit Differentiation – when variables appear in both base and exponent, take logarithms before differentiating.

Let $u = y^x$, $v = x^y$, $w = x^x$. Then $u + v + w = a^b$ (constant).

Step 1: Differentiate $u = y^x$

Take $\log u = x \log y$. Differentiate:

$\frac{1}{u} \frac{du}{dx} = \log y + \frac{x}{y} \frac{dy}{dx}$

So $\frac{du}{dx} = y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right)$.

Step 2: Differentiate $v = x^y$

Take $\log v = y \log x$. Differentiate:

$\frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} \log x + \frac{y}{x}$

So $\frac{dv}{dx} = x^y \left( \frac{dy}{dx} \log x + \frac{y}{x} \right)$.

Step 3: Differentiate $w = x^x$

Take $\log w = x \log x$. Differentiate:

$\frac{1}{w} \frac{dw}{dx} = \log x + 1$

So $\frac{dw}{dx} = x^x (1 + \log x)$. …

Why logarithmic differentiation?

When you see $y^x$, $x^y$, or $x^x$, the variable is in both the base and the exponent. The standard power rule ($\frac{d}{dx} x^n = n x^{n-1}$) only works when the exponent is a constant. The exponential rule ($\frac{d}{dx} a^x = a^x \log a$) only works when the base is constant. Here, neither is constant — so we need a different tool.

The trick: take natural logs first, then differentiate implicitly. This converts the variable exponent into a product, which we can handle with the product rule.

Step-by-step solution

1. Write the given equation

Here $a$ and $b$ are constants, so the right-hand side is a constant. That means its derivative is $0$.

2. Differentiate term by term — start with $y^x$

Let $u = y^x$. Take $\log$ on both sides:

Differentiate implicitly with respect to $x$ (remember $y$ is a function of $x$):

So

3. Differentiate $x^y$

Let $v = x^y$. Take $\log$:

Differentiate:

Thus

4. Differentiate $x^x$

Let $w = x^x$. Take $\log$:

Differentiate:

So

5. Combine the derivatives

The derivative of the whole left side equals the derivative of the constant right side ($0$):

6. Collect terms with $\frac{dy}{dx}$ …

Method: Combining Implicit and Logarithmic Differentiation for a Sum of Variable-Exponent Terms

Use this method when an equation contains several separate terms, each of the form (variable)^(variable) — like $y^x$, $x^y$, or $x^x$ — added together and set equal to a constant (or to each other).

Steps

Step 1: Differentiate each variable-exponent term separately using logarithmic differentiation

For a term like $u = y^x$, take $\log u = x\log y$ and differentiate implicitly (remembering $y = y(x)$):

Repeat this same process independently for every other variable-exponent term in the equation (e.g. $x^y$), and use the ordinary product/chain rule for any term where the exponent or base is fixed (e.g. $x^x$ still needs the log trick since both vary, but a term like $y^2$ would just need the plain chain rule).

Step 2: Differentiate the constant (or other) side of the equation

If the right-hand side is a constant (like $a^b$ with fixed $a,b$), its derivative is $0$.

Step 3: Add up the derivatives of every term to reconstruct the full differentiated equation …

Common Mistakes

Mistake 1: Differentiating $y^x$ as if only the exponent (or only the base) varies

Why it's wrong: Because $y^x$ has a variable base ($y$, which itself depends on $x$) and a variable exponent ($x$), its derivative genuinely has two separate contributions — one from the exponent ($\log y$) and one from the base changing ($\frac{x}{y}\frac{dy}{dx}$). Dropping either term (most often the $\frac{x}{y}\frac{dy}{dx}$ piece, since it's easy to forget $y$ is implicit) gives an incomplete derivative. Correct approach: always log-differentiate $y^x$ fully: $\frac{d}{dx}(y^x) = y^x\left(\log y + \frac{x}{y}\frac{dy}{dx}\right)$.

Mistake 2: Forgetting that the right-hand side, $a^b$, is a constant and differentiates to $0$

Why it's wrong: $a$ and $b$ are both stated to be fixed positive constants, so $a^b$ is just a fixed number — some students, seeing variables in the exponent elsewhere in the problem, mistakenly try to differentiate $a^b$ as though $a$ or $b$ depended on $x$. Correct approach: recognize $a^b$ as a plain constant and set its derivative to $0$ before proceeding. …