Implicit differentiation

Implicit differentiation is the process of deriving an equation without isolating y. It is used generally when it is difficult or impossible to solve for y. This is done by simply taking the derivative of every term in the equation (${\frac{dy}{dx}}$).

Examples

$x^2+y^2=12$

$\frac d{dx} (x^2 + y^2) = \frac d{dx}12$

$2x + 2y\frac{dy}{dx} = 0$

Note that $(y^2)'=y'\frac d{dx}y^2=2y\frac{dy}{dx}$ because of the chain rule. Also notice that the constant rule is not being applied to ${\textstyle y^2}$, because y is a function of x, that is, x affects y. From here, ${\frac{dy}{dx}}$ can be solved for as if it were a variable:

$2y {dy\over dx} = -2x,$

${dy\over dx} = -{2x\over 2y}$

${dy\over dx} = -{x\over y}$

This usually results in an answer that has both x and y in the formula.

Another example:

$\ln(xy) = -3y$

$\ln'(xy) = \frac d{dx}(-3y)$

Chain Rule:

$\frac {(xy)'}{xy} = \frac{dy}{dx} (-3)$

Use the product rule:

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

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

$\frac {dy}{dx} x + \frac{dy}{dx} 3xy = -y$

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

$\frac {dy}{dx} = -\frac y{x + 3xy}$

Second Derivative Examples

In order to take a second derivative ${d^2y\over (dx)^2}$ ($f''$) while differentiating implicitly, you take a derivative of the derivative:

Since ${dy\over dx} = -{x\over y}$
we use the quotient rule on x and y to find $f''$.

${d^2y\over (dx)^2} = {y(-1)-(-x){dy\over dx}) \over y^2}$

Then we can plug in for ${\frac{dy}{dx}}$, since we have that from above,

${d^2y\over (dx)^2} = {-y+ x{-x\over y}\over y^2}$

${d^2y\over (dx)^2} = {-y{-x^2\over y}\over y^2}$

${d^2y\over (dx)^2} = {-(x^2+y^2)\over y^3}$

Now, substituting the original equation $x^2+y^2=12$

${d^2y\over (dx)^2} = {-12\over y^3}$