Implicit Differentiation Math Example 2

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Example 2

hard
Find dydx\frac{dy}{dx} for x3+y3=6xyx^3 + y^3 = 6xy (folium of Descartes).

Solution

  1. 1
    Differentiate both sides with respect to xx:
  2. 2
    3x2+3y2dydx=6y+6xdydx3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx} (product rule on 6xy6xy).
  3. 3
    Collect dydx\frac{dy}{dx} terms: 3y2dydx−6xdydx=6y−3x23y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2.
  4. 4
    Factor: dydx(3y2−6x)=6y−3x2\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2.
  5. 5
    dydx=6y−3x23y2−6x=2y−x2y2−2x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.

Answer

dydx=2y−x2y2−2x\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}
The right side requires the product rule on 6xy6xy. After differentiating, move all dydx\frac{dy}{dx} terms to one side, factor out dydx\frac{dy}{dx}, and divide.

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 →

More Implicit Differentiation Examples

Example 1 easy

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

Example 3 medium

Find [formula] for the circle [formula].

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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