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Mathematics · Ch 5 — Continuity and Differentiability

Derivatives of Implicit Functions

5.3.2

Derivatives of Implicit Functions

Concept First: Explicit vs. Implicit Functions

So far you have differentiated functions in the form y=f(x)y = f(x) — an explicit function, where yy is given directly in terms of xx (e.g. y=x2+3xy = x^2 + 3x or y=sin⁡xy = \sin x). But many relations are not in this solved form. Consider:

  1. x−y−π=0x - y - \pi = 0
  2. x+sin⁡(xy)−y=0x + \sin(xy) - y = 0

The first solves easily to y=x−πy = x - \pi. In the second, solving for yy is not straightforward, yet yy still depends on xx — the relation just hides it. When yy is not isolated we call yy an implicit function of xx. The goal here is to find dydx\frac{dy}{dx} directly from such a relation, without solving for yy.


Differentiating Implicit Functions: The Core Idea

Differentiate every term of the relation with respect to xx, treating yy as a function of xx — so apply the chain rule to any term involving yy. For example:

ddx(y2)=2y⋅dydx\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}

ddx(sin⁡y)=cos⁡y⋅dydx\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}

ddx(xy)=x⋅dydx+y⋅1(product rule)\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1 \quad \text{(product rule)}

After differentiating every term, solve the resulting equation algebraically for dydx\frac{dy}{dx}.

Watch out

A Common Pitfall …