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What is Calculus I

Implicit
Differentiation?

Finding $\frac{dy}{dx}$ when $y$ is tangled up with $x$ in an equation and cannot be isolated — differentiate both sides and solve for $\frac{dy}{dx}$.

Simple Definition

Most functions you've seen are written explicitly: $y = x^2 + 3$. You can directly differentiate the right side. But some relationships between $x$ and $y$ can't be written that way — they're implicit: $x^2 + y^2 = 25$ (a circle). There's no clean way to solve for $y$ alone without square roots and sign choices.

Implicit differentiation solves this. Instead of isolating $y$ first, you differentiate both sides of the equation with respect to $x$ — treating $y$ as an unknown function of $x$. Every time you differentiate a $y$ term, you multiply by $\frac{dy}{dx}$ (chain rule!). Then you solve algebraically for $\frac{dy}{dx}$.

The result: you can find the slope of the tangent to any curve — circles, ellipses, Folium of Descartes — not just explicit functions.

Intuition

⭕

A circle is not a function

The circle $x^2 + y^2 = 25$ fails the vertical line test — it's not a function. But locally, near any point (except the very top and bottom), it behaves like a function. Implicit differentiation captures that local behaviour — it gives you the slope $\frac{dy}{dx}$ at any specific point on the circle, without forcing you to separate the circle into two halves.

🔗

y is secretly a function of x

When we write $x^2 + y^2 = 25$ and differentiate, we're assuming that locally, $y$ is some differentiable function of $x$ — even if we can't write it down explicitly. This assumption (the Implicit Function Theorem) is what makes the whole technique valid. We treat $y = y(x)$ and apply the chain rule.

The Key Rule

Chain Rule Applied to y

$$\frac{d}{dx}[y^n] = n\,y^{n-1}\cdot\frac{dy}{dx}$$

This is the heart of implicit differentiation. Every time you differentiate any expression involving $y$, the chain rule forces you to multiply by $\frac{dy}{dx}$. Terms involving only $x$ differentiate normally.

Step-by-Step Method

Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$:

Differentiate both sides with respect to $x$

$\dfrac{d}{dx}[x^2] + \dfrac{d}{dx}[y^2] = \dfrac{d}{dx}[25]$

Apply chain rule to any $y$ term

$2x + 2y\,\dfrac{dy}{dx} = 0$

Isolate $\frac{dy}{dx}$

$2y\,\dfrac{dy}{dx} = -2x \implies \dfrac{dy}{dx} = -\dfrac{x}{y}$

Substitute a point to get the slope

At $(3, 4)$: $\dfrac{dy}{dx} = -\dfrac{3}{4}$. Tangent line slope = −3/4.

Visual — Tangent Lines on a Circle

Move mouse over the circle — tangent line updates using dy/dx = −x/y from implicit differentiation

Worked Example — Harder Case

Find dy/dx for x³ + y³ = 6xy (Folium of Descartes)

Differentiate

$3x^2 + 3y^2\,\dfrac{dy}{dx} = 6y + 6x\,\dfrac{dy}{dx}$

Collect dy/dx

$3y^2\,\dfrac{dy}{dx} - 6x\,\dfrac{dy}{dx} = 6y - 3x^2$

Factor

$\dfrac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$

Solve

$\dfrac{dy}{dx} = \dfrac{6y - 3x^2}{3y^2 - 6x} = \dfrac{2y - x^2}{y^2 - 2x}$

Quick Examples

Example 1 — Ellipse

$\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$

Differentiate: $\dfrac{2x}{9} + \dfrac{2y}{4}\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = -\dfrac{4x}{9y}$

Example 2 — Mixed terms

$xy + y^2 = 10$

Product rule on $xy$:
$y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{-y}{x+2y}$

Example 3 — With trig

$\sin(xy) = x$

Chain + product rule:
$\cos(xy)\left(y + x\dfrac{dy}{dx}\right) = 1$
Solve for $\dfrac{dy}{dx}$

Example 4 — Tangent line

$x^2 + y^2 = 25$ at $(0,5)$

$\dfrac{dy}{dx} = -\dfrac{x}{y} = -\dfrac{0}{5} = 0$
Horizontal tangent at the top of the circle ✓

Common Mistakes

✗ Forgetting dy/dx when differentiating y terms

This is the most common error. $\frac{d}{dx}[y^2] = 2y$ is wrong. The correct result is $2y\frac{dy}{dx}$. Every $y$ term, when differentiated with respect to $x$, must carry a $\frac{dy}{dx}$ factor by the chain rule.

✗ Differentiating mixed terms incorrectly

For a product like $xy$, you must use the product rule: $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$. Students often write just $y$ or just $x\frac{dy}{dx}$ — both are wrong. The product rule gives both terms.

✗ Leaving dy/dx on both sides without solving

After differentiating, collect all $\frac{dy}{dx}$ terms on one side, factor it out, and divide. The final answer must be a single expression for $\frac{dy}{dx}$, possibly in terms of both $x$ and $y$ — that's perfectly fine.

Related Concepts

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