Calculus/Implicit differentiation

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

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,

$y^{\prime }=-\frac{x}{y}$

Implicit differentiation

Using the same equation

$x^2+y^2=1$

First, differentiate the individual terms of the equation:

$2x+2y{y}^{\prime }=0$

Separate the variables:

$2y{y}^{\prime }=-2x$

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

$y^{\prime }=-\frac{2x}{2y}$

$y^{\prime }=-\frac{x}{y}$

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) :

$2x+y+x{y}^{\prime }+y{y}^{\prime }=0$

Isolate terms with y':

$x{y}^{\prime }+y{y}^{\prime }=-2x-y$

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

$y^{\prime }=\frac{-2x-y}{x+y}$