Q.If $y = x^y$, then $\dfrac{dy}{dx}$ is : (A) $x^y (\log x + 1)$ (B) $\dfrac{y^2}{x(1 + y \log x)}$ (C) $x^y (\log x - 1)$ (D) $\dfrac{y^2}{x(1 - y \log x)}$

Implicit Differentiation: The Intuition

You already know how to differentiate $y = x^2 + 3x$ — just apply the power rule and get $\frac{dy}{dx} = 2x + 3$. That's explicit differentiation: $y$ is written directly in terms of $x$, so the derivative falls out cleanly.

But what if you're given something like $x^2 + y^2 = 25$? Here $y$ is not isolated. You could solve for $y$ (getting $y = \pm \sqrt{25 - x^2}$) and then differentiate — but that's messy, and you'd have to handle the $\pm$ separately. Worse, try solving $y^3 + xy + x^3 = 1$ for $y$. It's impossible by elementary means.

Implicit differentiation is the trick that lets you find $\frac{dy}{dx}$ without isolating $y$ first. The core idea is simple: treat $y$ as an unknown function of $x$, and differentiate both sides of the equation with respect to $x$, using the chain rule whenever you hit a $y$.

The Precise Statement

Given an equation relating $x$ and $y$ (like $F(x, y) = 0$), differentiate every term with respect to $x$, remembering that $y$ is a function of $x$. Whenever you differentiate a term containing $y$, apply the chain rule:

Then solve the resulting equation for $\frac{dy}{dx}$.

Worked Example: $x^2 + y^2 = 25$

Step 1: Differentiate both sides with respect to $x$.

• $\frac{d}{dx}(x^2) = 2x$
• $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ (chain rule: derivative of $y^2$ is $2y$, times derivative of $y$)
• $\frac{d}{dx}(25) = 0$

So we get:

Step 2: Solve for $\frac{dy}{dx}$.

That's it. The derivative is expressed in terms of both $x$ and $y$ — which is natural, because the slope of the circle at a point depends on where you are.

Why It Works

The chain rule is the engine. When you write $y^2$, you're really writing $[y(x)]^2$ — a function of a function. Differentiating it requires the chain rule, and that's exactly what produces the $\frac{dy}{dx}$ factor. Every term with $y$ contributes one such factor; terms with only $x$ differentiate normally.

Another Example: $y^3 + xy + x^3 = 1$

Differentiate term by term:

• $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$ …