Implicit Differentiation Math Example 3

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Example 3

medium
Find dydx\frac{dy}{dx} for the circle x2+y2=25x^2 + y^2 = 25.

Solution

  1. 1
    Differentiate both sides with respect to xx: ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25).
  2. 2
    Apply the chain rule to y2y^2: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.
  3. 3
    Solve for dydx\frac{dy}{dx}: 2ydydx=โˆ’2x2y\frac{dy}{dx} = -2x, so dydx=โˆ’2x2y=โˆ’xy\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y} (where yโ‰ 0y \neq 0).

Answer

dydx=โˆ’xy\frac{dy}{dx} = -\frac{x}{y}
The circle x2+y2=25x^2 + y^2 = 25 cannot be written as a single function y=f(x)y = f(x), so we differentiate implicitly. The chain rule on y2y^2 produces 2ydydx2y\frac{dy}{dx} because yy is a function of xx. The negative sign reflects the fact that as xx increases along the circle, yy decreases (in the upper half).

About Implicit Differentiation

Finding dydx\frac{dy}{dx} when yy is defined implicitly by an equation like F(x,y)=0F(x, y) = 0, by differentiating both sides and solving for dydx\frac{dy}{dx}.

Learn more about Implicit Differentiation โ†’

More Implicit Differentiation Examples

Example 1 easy

Find [formula] for the circle [formula] and evaluate it at the point [formula].

Example 2 hard

Find [formula] for [formula] (folium of Descartes).

Example 4 easy

Find [formula] for [formula].

Example 5 medium

Find the equation of the tangent line to [formula] at the point [formula].

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