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Implicit differentiation

Implicit differentiation involves differentiating implicit functions. An implicit function is an implicit relation between variables. Differentiating an implicit function leads to differentiate the independent variable with respect to the dependent variable. It's basically finding the derivative using the notation dy/dx.
Two methods can be used:
1) You explicit the function

**Example 1**

Find the derivative of 3xy = 2
Let's explicit the function: y = 2/3x
Let's calculate the derivative: dy/dx = d/dx(2/3x)
= -2(3x)'/(3x)²
= -6./9x²
= -2/3x²
2) If expliciting is not possible, you make transformations in order to find the derivative.

The rules and formulas used to calculate the derivative of different forms of functions apply in the calculations of the derivative of an implicit function.

Since an implicit function is a relationship between the independent and the dependent variable the the application of the derivative rules might seem odd. Let's familiarize ourselves with the derivatives of some expressions where the derivative rules are applied.
**Example 2**. Let's y be a function of x find the derivative of y³ with respect to x.
Let's u = y³. we have two functions: u and y. U is a function of y and y is a function of x. U is a composite function. The chain rule has to be applied in order to find the derivative. The formula to apply here is du/dx = du/dy.dy/dx
du/dx = d(y³)/dy.dy/dx
= 3y²dy/dx.
**Example 3** Find the derivative of u = 2x²y
du/dx = d(2x²y)
Let's apply the constant rule
du/dx = 2d(x²y)
Let's apply the product rule:
du/dx = 2[d/dx(x²)y+x² dy/dx]
= :2(2xy+x²dy/dx)
du/dx = 4xy+2x²dy/dx
**Example 4** Find the derivative of 3y³+x²y = x-3
Let's differentiate both sides:
d/dx(3y³+x²y) = d/dx(x-3)
3y²dy/dx+2xy+x²dy/dx = 1
3y²dy/dx+x²dy/dx = 1-2xy
(3y²+x²)dy/dx = 1-2xy
dy/dx = 1-2xy/3y²+x²

**Practice**. Find the derivatives of the implicit functions:
1) x²+y² = 15
2) 3y²-siny = x²
3) x²+2xy-y = 2
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