Calculus/Implicit differentiation

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Implicit differentiation

Implicit differentiation takes a relation and turns it into a rectangular regular equation.

[edit] Explicit differentiation

For example, to differentiate a function explicitly,

x^2 + y^2 = 1\,

First we can separate variables to get

y^2 = 1 - x^2\,

Taking the square root of both sides we get a function of y:

y = \pm \sqrt{1-x^2}\,

We can rewrite this as a fractional power as

y = \pm (1-x^2)^{\frac{1}{2}}\,

Using the chain rule and simplifying we get,

y' = -\frac{x}{y}

[edit] Implicit differentiation

Using the same equation

x^2 + y^2 = 1\,

First, differentiate the individual terms of the equation:

2x + 2yy' = 0\,

Separate the variables:

2yy' = -2x\,

Divide both sides by 2y\,, and simplify to get the same result as above:

y' = -\frac{2x}{2y}
y' = -\frac{x}{y}

[edit] Uses

Implicit differentiation is useful when differentiating an equation that cannot be explicitly differentiated because it is impossible to isolate variables.

For example, consider the equation,

x^2 + xy + y^2 = 16\,

Differentiate both sides of the equation (remember to use the product rule on the term xy) :

2x + y + xy' + 2yy' = 0\,

Isolate terms with y':

xy' + 2yy' = -2x - y\,

Factor out a y' and divide both sides by the other term:

y' = \frac{-2x-y}{x+2y}

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