Q.If $y^x = e^{y - x}$, prove that $\dfrac{dy}{dx} = \dfrac{(1 + \log y)^2}{\log y}$.

Concept: Implicit Differentiation — take logs to handle the variable exponent, then differentiate both sides with respect to $x$.

Take natural logarithms on both sides:

Differentiate implicitly with respect to $x$:

Collect terms with $\frac{dy}{dx}$:

From $x \log y = y - x$, we have $x(1 + \log y) = y$, so $\frac{x}{y} = \frac{1}{1 + \log y}$. Substitute: …

We start with the equation

The presence of $x$ in the exponent of $y$ and the exponential on the right makes direct differentiation messy. The natural move is to take the logarithm of both sides — this brings the exponent down and turns the product into a sum, making implicit differentiation straightforward.

Step 1: Take the natural logarithm of both sides.

Since $y > 0$ (otherwise $y^x$ is not defined for real numbers), we can safely write:

Using logarithm properties:

This is now a simple implicit relation between $x$ and $y$.

Step 2: Differentiate both sides with respect to $x$.

Remember that $y$ is a function of $x$, so $\log y$ differentiates to $\frac{1}{y} \cdot \frac{dy}{dx}$.

On the left, we have a product $x \cdot \log y$:

On the right, $y - x$ differentiates to:

So the differentiated equation is:

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

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

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

Thus:

Step 4: Eliminate $x$ using the original relation.

From $x \log y = y - x$, we can solve for $x$:

Substitute this into $x - y$: …

Method: Logarithmic Differentiation for an Implicit Relation with a Variable Exponent

Use this when the equation has a variable raised to a power that is also a variable (like $y^x$), making direct differentiation impossible — neither the power rule ($x^n$, constant exponent) nor the exponential rule ($a^x$, constant base) applies on its own.

Steps

Step 1: Take the natural logarithm of both sides

For $y^x = e^{y-x}$:

using $\log(a^b)=b\log a$ and $\log(e^k)=k$. This brings the exponent down and turns the equation into one built from products and sums, which standard differentiation rules can handle.

Step 2: Differentiate implicitly with respect to $x$

The left side, $x\log y$, is a product of $x$ and $\log y$ (where $\log y$ needs the chain rule since $y=y(x)$):

The right side differentiates directly:

Step 3: Collect $\dfrac{dy}{dx}$ terms and factor

Step 4: Use the original relation to eliminate $x$ …

Common Mistakes

Mistake 1: Trying to differentiate $y^x$ directly without taking logs

$y^x$ has both a variable base and a variable exponent, so neither the power rule nor the exponential rule alone applies. Students who skip logarithmic differentiation often apply the wrong rule (treating it like $x^n$ or like $a^x$) and get a completely incorrect derivative. Why it's wrong: only after taking $\log$ of both sides does $y^x=e^{y-x}$ become the tractable equation $x\log y=y-x$. Correct approach: always take logs first whenever both the base and the exponent involve the variable.

Mistake 2: Forgetting the chain rule on $\log y$

When differentiating $x\log y$ via the product rule, a frequent slip is writing $\frac{d}{dx}(\log y)=\frac{1}{y}$ instead of $\frac{1}{y}\frac{dy}{dx}$. Why it's wrong: $y$ depends on $x$, so $\log y$ needs the chain rule just like any other $y$-term. Correct approach: always attach $\frac{dy}{dx}$ to the derivative of $\log y$. …