Q.Find $\frac{dy}{dx}$ in the following: $y^x = x^y$

The key idea is implicit differentiation combined with logarithmic differentiation, since the variable appears in both the base and the exponent.

Take the natural logarithm of both sides:

Differentiate implicitly with respect to $x$ (using the product rule on each side):

Collect terms with $\frac{dy}{dx}$ on one side:

Factor and solve: …

Why implicit differentiation?

The equation $y^x = x^y$ is not solved for $y$ in terms of $x$ — and trying to solve it explicitly is messy. Both $x$ and $y$ appear in the exponent, so we need a technique that handles variables in both base and exponent. That technique is logarithmic differentiation: take the natural log of both sides, which brings the exponents down as coefficients, then differentiate implicitly.

Step-by-step solution

1. Take the natural logarithm of both sides

Using the power rule for logs: $\log(a^b) = b \log a$, we get

Now the equation is in a form we can differentiate.

2. Differentiate both sides with respect to $x$

Remember that $y$ is a function of $x$, so whenever we differentiate a term involving $y$, we must multiply by $\frac{dy}{dx}$ (the chain rule).

Left side: differentiate $x \log y$ using the product rule.

So the left side becomes

Right side: differentiate $y \log x$ using the product rule.

So the right side becomes

3. Collect terms with $\frac{dy}{dx}$

We now have

Bring the $\frac{dy}{dx}$ terms to one side:

Factor out $\frac{dy}{dx}$:

4. Solve for $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x}$ …

Method: Taking Logs of Both Sides When Both Sides Are Variable-Exponent Expressions

When an equation has the form $[\text{expr}_1]^{[\text{expr}_2]} = [\text{expr}_3]^{[\text{expr}_4]}$ (a power on each side, not a sum), take the logarithm of the entire equation once, not term by term — the log turns each power into a simple product.

Steps

Step 1: Apply $\log$ to both sides and bring down the exponents

Both sides are now products of an ordinary variable and a logarithm — much easier to differentiate.

Step 2: Differentiate both sides implicitly using the product rule

On each side, one factor is a "plain" variable and the other is a logarithm of the other variable. Differentiate $\log y$ using the chain rule (it hides $y(x)$):

Step 3: Collect every $\dfrac{dy}{dx}$ term on one side

Move all terms containing $\dfrac{dy}{dx}$ to the left and everything else to the right, then factor: …

Common Mistakes

Mistake 1: Assuming $y^x = x^y$ forces $y = x$, and simplifying the answer prematurely

Why it's wrong: $y=x$ is only one family of solutions to this equation — pairs like $(2,4)$ and $(4,2)$ also satisfy $y^x=x^y$ (since $2^4=4^2=16$) without $x=y$. Substituting $y=x$ before differentiating throws away the general relationship the problem is actually asking about. Correct approach: keep $x$ and $y$ as independent-looking variables linked only by the given equation, and differentiate implicitly without assuming any special relationship between them.

Mistake 2: Taking logs of $y^x$ and $x^y$ as if they were separate additive terms …