Difference between revisions of "Implicit differentiation"

Revision as of 10:12, 29 October 2008

Implicit differentiation is differentiating both sides of an implicit equation with respect to one of the variables. The dependent variable is treated as a function of the independent variable and is differentiated with the chain rule.

$x^3 + xy^2 + x^2y + y^3 + sin(xy) = 2xy$

$3x^2 + (1y^2 + 2xy \frac{dy}{dx}) + (2xy + x^2 \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} + (cos(xy) (1y + 1x \frac{dy}{dx}) = 2 (1y + 1x \frac{dy}{dx})$

$\frac{dy}{dx} (2xy + x^2 + 3y^2 + xcos(xy) - 2x) = -3x^2 - y^2 - 2xy - ycos(xy) + 2y$

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2 + ycos(xy) - 2y}{x^2 + 2xy + 3y^2 + xcos(xy) - 2x}$

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 == Example ==
-{{incomplete}}
+<math>x^3 + xy^2 + x^2y + y^3 + sin(xy) = 2xy</math>
+<math>3x^2 + (1y^2 + 2xy \frac{dy}{dx}) + (2xy + x^2 \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} + (cos(xy) (1y + 1x \frac{dy}{dx}) = 2 (1y + 1x \frac{dy}{dx})</math>
+<math>\frac{dy}{dx} (2xy + x^2 + 3y^2 + xcos(xy) - 2x) = -3x^2 - y^2 - 2xy - ycos(xy) + 2y</math>
+<math>\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2 + ycos(xy) - 2y}{x^2 + 2xy + 3y^2 + xcos(xy) - 2x}</math>
 {{stub}}