Q.Find $\frac{dy}{dx}$ in the following: $x^y + y^x = 1$

Concept: Implicit Differentiation — when the variables are mixed in exponents, we take logs of each term separately before differentiating.

Take the equation:

Let $u = x^y$ and $v = y^x$. Then $\frac{du}{dx} + \frac{dv}{dx} = 0$.

For $u = x^y$, take $\log u = y \log x$. Differentiate:

For $v = y^x$, take $\log v = x \log y$. Differentiate:

Add them and set to zero: …

We have an equation where the variable appears in both the base and the exponent: $x^y$ and $y^x$. 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 a constant. Here, both are moving — so neither rule applies directly.

The trick is to take the natural logarithm of each term separately, which brings the exponent down as a coefficient, turning the problem into one of implicit differentiation.

1. Rewrite each term using logarithms. Let $u = x^y$ and $v = y^x$. Then $\log u = y \log x$ and $\log v = x \log y$. So $u = e^{y \log x}$ and $v = e^{x \log y}$. The equation becomes:

2. Differentiate both sides with respect to $x$. Remember $y$ is a function of $x$, so every time we differentiate a $y$, we multiply by $\frac{dy}{dx}$ (implicit differentiation). For the first term:

Now $\frac{d}{dx}(y \log x) = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x}$.

So the derivative of the first term is:

For the second term:

Now $\frac{d}{dx}(x \log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}$.

So the derivative of the second term is:

The derivative of the right-hand side (1) is 0.

3. Assemble the differentiated equation:

4. Collect all terms containing $\frac{dy}{dx}$ on one side. Expand:

Simplify $x^y \cdot \frac{y}{x} = y x^{y-1}$ and $y^x \cdot \frac{x}{y} = x y^{x-1}$.

So: …

Method: Implicit + Logarithmic Differentiation for $f(x)^{g(x)}$ Terms Buried in an Equation

When an equation mixes ordinary terms with $x$-in-the-exponent terms (like $x^y$ or $y^x$) and isn't solved for $y$, combine implicit differentiation (because $y$ is tangled with $x$) with logarithmic differentiation (because the exponent itself varies).

Steps

Step 1: Isolate each variable-exponent term and take its own logarithm

For $u = x^y$: $\log u = y\log x$. For $v = y^x$: $\log v = x\log y$. Do this term by term — never take one big logarithm of the whole equation if there are added or subtracted terms.

Step 2: Differentiate each log implicitly, remembering $y=y(x)$

Use the product rule on the right-hand side of each, and every time you differentiate a $y$, attach a factor of $\dfrac{dy}{dx}$:

Step 3: Multiply back by $u$ and $v$, then combine per the original equation …

Common Mistakes

Mistake 1: Treating $x^y$ with the ordinary power rule

Why it's wrong: the power rule $\frac{d}{dx}x^n = nx^{n-1}$ only holds when the exponent $n$ is a genuine constant. Here the exponent is $y$, which is itself a function of $x$ — applying the power rule as if $y$ were fixed silently drops an entire term (the one coming from $y$ changing with $x$). Correct approach: take $\log u = y\log x$ first, so both the varying base and varying exponent are properly accounted for.

Mistake 2: Losing track of which factor was actually differentiated once terms like $\dfrac{y}{x}$ appear …