Q.Find $\dfrac{dy}{dx}$ from the following parametric equations i. $x = at$, $y = \dfrac{a}{t}$
ii. $x = t\cdot\log t$, $y = \dfrac{\log t}{t}$
iii. $x = \dfrac{a(1-t^2)}{1+t^2}$, $y = \dfrac{2bt}{1+t^2}$

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:

• $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$
• $\frac{d}{dx}(xy)$: use product rule — $x$ times $y$ gives $1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$
• $\frac{d}{dx}(x^3) = 3x^2$
• $\frac{d}{dx}(1) = 0$

Put it together:

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

Solve:

No solving for $y$ needed — just algebra after differentiation.

When to Use Implicit Differentiation

Use it whenever:

• $y$ is difficult or impossible to isolate
• The equation involves products or compositions of $x$ and $y$ (like $xy$, $e^{xy}$, $\sin(xy)$)
• You need the derivative at a specific point without solving for $y$ explicitly

Summary

Implicit differentiation is just the chain rule applied to an equation. Differentiate both sides with respect to $x$, treat $y$ as $y(x)$, collect $\frac{dy}{dx}$ terms, and solve. It's a mechanical process — once you practice it, it becomes as automatic as explicit differentiation.