Implicit Differentiation

Differentiating equations where y is not isolated β€” treating y as a function of x and applying the chain rule to every term.

Definition

Implicit differentiation finds dydx\frac{dy}{dx} when yy is defined implicitly by an equation in xx and yy, rather than explicitly as y=f(x)y = f(x).

The key rule: treat yy as a function of xx, and apply the chain rule whenever you differentiate a term involving yy:

ddx[yn]=nynβˆ’1β‹…dydx\frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx}

Procedure:

  1. Differentiate both sides with respect to xx.
  2. Apply the chain rule to every yy term (multiply by dydx\frac{dy}{dx}).
  3. Collect all dydx\frac{dy}{dx} terms on one side.
  4. Solve for dydx\frac{dy}{dx}.
Circle: xΒ² + yΒ² = rΒ²

Differentiate x2+y2=r2x^2 + y^2 = r^2 with respect to xx (rr is constant):

2x+2ydydx=0β€…β€ŠβŸΉβ€…β€Šdydx=βˆ’xy2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

At the point (3,4)(3, 4) on a circle of radius 55: dydx=βˆ’34\frac{dy}{dx} = -\frac{3}{4}.

This matches the geometric fact that the tangent to a circle is perpendicular to the radius (slope of radius =4/3= 4/3, perpendicular slope =βˆ’3/4= -3/4).

Try it

Find dydx\frac{dy}{dx} for the ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1. Find the slope at (0,2)(0, 2).

Solution

Differentiate: 2x9+2y4dydx=0β€…β€ŠβŸΉβ€…β€Šdydx=βˆ’4x9y\frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y}.

At (0,2)(0, 2): dydx=βˆ’4(0)9(2)=0\frac{dy}{dx} = -\frac{4(0)}{9(2)} = 0. The tangent is horizontal β€” the ellipse is at its top.

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