Q.If $x = e^{x/y}$, prove that $\dfrac{dy}{dx} = \dfrac{x - y}{x \log x}$.

Concept: Implicit Differentiation — we differentiate the given equation treating $y$ as a function of $x$, then solve for $\frac{dy}{dx}$.

Given $x = e^{x/y}$, take the natural logarithm on both sides:

Differentiate both sides with respect to $x$ (using the product rule on the left):

Solve for $\frac{dy}{dx}$: …

Why implicit differentiation?

The equation $x = e^{x/y}$ ties $x$ and $y$ together in a way that can't be easily solved for $y$ as a function of $x$ (at least not in a simple closed form). But we don't need to — we can differentiate both sides with respect to $x$, treating $y$ as a function of $x$, and then solve for $\frac{dy}{dx}$. That's the essence of implicit differentiation.

A natural first step: take the natural logarithm on both sides. This simplifies the exponential and brings the variables into a more manageable form.

1. Take the natural log Given $x = e^{x/y}$, take $\log$ on both sides:

This is much cleaner. Now $y$ appears in the denominator, but we can work with it.

2. Differentiate implicitly Differentiate both sides with respect to $x$. Remember $y$ is a function of $x$, so use the chain rule on the right-hand side:

Left side: $\frac{1}{x}$.

Right side: use the quotient rule (or product rule — treat as $x \cdot y^{-1}$). Let's do quotient rule:

So we have:

3. Solve for $\frac{dy}{dx}$ Multiply both sides by $y^2$:

Rearrange to isolate the derivative term:

Factor $y$ on the right:

So:

4. Use the original relation to rewrite in terms of $x$ only We have $\log x = \frac{x}{y}$, so $y = \frac{x}{\log x}$. Substitute this into the expression:

Simplify numerator: …

Method: Taking Logarithms First to Simplify an Exponential Implicit Relation

Use this method whenever a variable appears in an exponent that itself involves both $x$ and $y$, such as $x=e^{x/y}$. Differentiating an exponential like this directly is messy; taking logarithms first almost always simplifies it.

Steps

Step 1: Take the natural logarithm of both sides

using $\log(e^k)=k$. This converts an exponential equation into a simple algebraic (rational) relation between $x$ and $y$ — much easier to differentiate.

Step 2: Clear any fraction by cross-multiplying, if convenient

This form avoids differentiating a quotient later.

Step 3: Differentiate implicitly with respect to $x$, using the product rule

$\frac{dy}{dx}\log x + y\cdot\frac{1}{x} = 1$ …

Common Mistakes

Mistake 1: Differentiating $e^{x/y}$ directly instead of taking logs first

Students sometimes try to apply the chain rule directly to $x=e^{x/y}$, differentiating the exponential with the quotient $x/y$ inside — this quickly becomes an unmanageable tangle of quotient-rule-inside-chain-rule terms. Why it's wrong: it's technically possible but extremely error-prone. Correct approach: take $\log$ of both sides first to turn the exponential relation into the much simpler $y\log x = x$.

Mistake 2: Forgetting the product rule on $y\log x$

After simplifying to $y\log x=x$, a common slip is writing $\frac{d}{dx}(y\log x)=\frac{y}{x}$ only, forgetting the $\frac{dy}{dx}\log x$ term from the product rule. Why it's wrong: $y$ is a function of $x$, so differentiating a product involving $y$ always needs both terms. Correct approach: apply the product rule fully: $\frac{dy}{dx}\log x + y\cdot\frac{1}{x}$. …