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

The key idea is Implicit Differentiation — since $y$ appears in both the base and the exponent, we take logs first to bring the powers down.

Take natural log on both sides:

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

Collect terms with $\frac{dy}{dx}$: …

We have an equation where the variable $y$ appears both as an exponent and inside a trigonometric function. Direct differentiation is impossible because the power rule and exponential rule don't apply when both base and exponent are functions of $x$. The standard technique here is logarithmic differentiation: take the natural logarithm of both sides, use log properties to bring down the exponent, then differentiate implicitly.

Let’s work through it step by step.

1. Take the natural log of both sides Start with $(\cos x)^y = (\cos y)^x$. Apply $\log$ to both sides:

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

2. Differentiate implicitly with respect to $x$ Both sides are products of functions of $x$ (remember $y$ is a function of $x$). Use the product rule on each side. Left side: differentiate $y \cdot \log(\cos x)$.
   • Derivative of $y$ is $\frac{dy}{dx}$ (call it $y'$).
   • Derivative of $\log(\cos x)$ is $\frac{1}{\cos x} \cdot (-\sin x) = -\tan x$. So by product rule:

Right side: differentiate $x \cdot \log(\cos y)$.

• Derivative of $x$ is $1$.
• Derivative of $\log(\cos y)$ is $\frac{1}{\cos y} \cdot (-\sin y) \cdot y' = -(\tan y) \, y'$ (chain rule). So by product rule:

3. Set the derivatives equal From step 1, the original equation after logs is an identity, so their derivatives are equal: …

Method: Logarithmic + Implicit Differentiation When the Variable Sits Inside a Trig Function AND an Exponent

This is the same $A^B=C^D$ pattern as taking logs of both sides, but here the bases are trigonometric functions of $x$ and $y$ — so differentiating the logs also needs the derivative of $\log(\cos\theta)$.

Steps

Step 1: Take $\log$ of both sides to remove the exponents

Step 2: Know the building-block derivative $\dfrac{d}{d\theta}\log(\cos\theta) = -\tan\theta$

This comes from $\dfrac{d}{d\theta}\log(\cos\theta) = \dfrac{1}{\cos\theta}\cdot(-\sin\theta) = -\tan\theta$. You'll need it twice: once for $\theta=x$ (plain), once for $\theta=y$ (where the chain rule adds a $\dfrac{dy}{dx}$).

Step 3: Differentiate both sides with the product rule

Left side ($y\cdot\log(\cos x)$): $\dfrac{dy}{dx}\log(\cos x) + y\cdot(-\tan x)$. …

Common Mistakes

Mistake 1: Forgetting that BOTH $\log(\cos x)$ and $\log(\cos y)$ contribute a negative sign

Why it's wrong: $\frac{d}{d\theta}\log(\cos\theta) = -\tan\theta$ for both $\theta=x$ and $\theta=y$; missing the minus sign on just one of the two occurrences (easy to do, since one differentiation is "plain" and the other carries an extra $\dfrac{dy}{dx}$ from the chain rule) throws off the sign of one whole side of the equation. Correct approach: write out $\frac{d}{dx}\log(\cos x) = -\tan x$ and $\frac{d}{dy}\log(\cos y) = -\tan y$ explicitly before substituting, so the sign is never in doubt.

Mistake 2: Forgetting the chain-rule factor of $\dfrac{dy}{dx}$ on $\log(\cos y)$ …