Implicit Differentiation

Functions are described explicitly when one variable can be written in terms of another variable. For example y = 2x³ - 5 describes the variable y in terms of the variable x and thus is described explicitly.

EXPLICIT: one variable is described in terms of one other variable.

y = 2x³ - 5

In other situations functions are described implicitly using the relationship between the two variables such as, x⁴ + y² = 1 - 5xy.

IMPLICIT: variables are NOT is described in terms of one other variable.

x⁴ + y² = 1 - 5xy

Functions described implicitly can be differentiated without being written explicitly. This type of differentiation is called Implicit Differentiation.

To differentiate implicitly both sides of the equation are differentiated with respect to x and the resulting equation solved for

\frac{\partial y}{\partial x}

.

Let's look at some examples:

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.

x² + y² = 36

Step 1: Differentiate the x-terms on both side of the equation with respect to x.

$\frac{\partial}{\partial x} (x^{2} + y^{2}) = \frac{\partial}{\partial x} 36$

Apply Sum Rule.

$\frac{\partial}{\partial x} x^{2} + \frac{\partial}{\partial x} y^{2} = \frac{\partial}{\partial x} 36$

Apply Power and Constant Rule.

$2 x + \frac{\partial}{\partial x} y^{2} = 0$

Step 2: Differentiate the y-terms.

Factor $\frac{\partial}{\partial x}$ into $\frac{\partial}{\partial y} \frac{\partial y}{\partial x}$ and take the derivative of $\frac{\partial}{\partial y} y^{2}$ normally resulting in the derivative times $\frac{\partial y}{\partial x}$ .

$2 x + \frac{\partial}{\partial x} y^{2} = 0$ Original

$\frac{\partial}{\partial x} y^{2} = \frac{\partial}{\partial y} y^{2} \frac{\partial y}{\partial x} = 2 y \frac{\partial y}{\partial x}$ Implicit Diff.

$2 x + 2 y \frac{\partial y}{\partial x} = 0$ Substitute

Step 3: Solve for $\frac{\partial y}{\partial x}$ and simplify.

Original

$2 x + 2 y \frac{\partial y}{\partial x} = 0$

Subtract 2x to both sides:

$2 y \frac{\partial y}{\partial x} = - 2 x$

Divide both sides by 2y.

$\frac{\partial y}{\partial x} = - \frac{2 x}{2 y}$

Simplify

$- \frac{x}{y}$

Example 1: 3y³ - x³ = 5xy

Step 1: Differentiate the x-terms on both side of the equation with respect to x.

$\frac{\partial}{\partial x} (3 y^{3} - x^{3}) = \frac{\partial}{\partial x} 5 x y$

Apply Difference Rule.

$\frac{\partial}{\partial x} 3 y^{3} - \frac{\partial}{\partial x} x^{3} = \frac{\partial}{\partial x} 5 x y$

Apply Constant Multiple Rule.

$3 \frac{\partial}{\partial x} y^{3} - \frac{\partial}{\partial x} x^{3} = 5 \frac{\partial}{\partial x} x y$

Apply Power Rule.

$3 \frac{\partial}{\partial x} y^{3} - 3 x^{2} = 5 \frac{\partial}{\partial x} x y$

Step 2: Differentiate the y-terms.

Factor $\frac{\partial}{\partial x}$ into $\frac{\partial}{\partial y} \frac{\partial y}{\partial x}$ and take the derivative of $\frac{\partial}{\partial y}$ normally resulting in the derivative times $\frac{\partial y}{\partial x}$ .

Original

$3 \frac{\partial}{\partial x} y^{3} - 3 x^{2} = 5 \frac{\partial}{\partial x} x y$

Implicitly differentiate $3 \frac{\partial}{\partial x} y^{3}$ using the power rule.

$3 \frac{\partial}{\partial x} y^{3} = 3 \frac{\partial}{\partial y} y^{3} \frac{\partial y}{\partial x} = 9 y^{2} \frac{\partial y}{\partial x}$

Implicitly differentiate $5 \frac{\partial}{\partial x} x y$ using the product rule.

$5 \frac{\partial}{\partial x} x y = 5 [(x \cdot 1) \frac{\partial y}{\partial x} + (y \cdot 1)]$

$5 x \frac{\partial y}{\partial x} + 5 y$

Substitute into the original equation:

$9 y^{2} \frac{\partial y}{\partial x} - 3 x^{2} = 5 x \frac{\partial y}{\partial x} + 5 y$

Step 3: Solve for $\frac{\partial y}{\partial x}$ and simplify.

Original

$9 y^{2} \frac{\partial y}{\partial x} - 3 x^{2} = 5 x \frac{\partial y}{\partial x} + 5 y$

Add 3x² to both sides:

$9 y^{2} \frac{\partial y}{\partial x} = 5 x \frac{\partial y}{\partial x} + 5 y + 3 x^{2}$

Subtract $5 x \frac{\partial y}{\partial x}$ from both sides.

$9 y^{2} \frac{\partial y}{\partial x} - 5 x \frac{\partial y}{\partial x} = 5 y + 3 x^{2}$

Factor out $\frac{\partial y}{\partial x}$ from the left-hand side:

$\frac{\partial y}{\partial x} (9 y^{2} - 5 x) = 5 y + 3 x^{2}$

Divide both sides by 9y² - 5x.

$\frac{\partial y}{\partial x} = \frac{5 y + 3 x^{2}}{9 y^{2} - 5 x}$

Example 2: e^y = 2x - y

Step 1: Differentiate the x-terms on both side of the equation with respect to x.

$\frac{\partial}{\partial x} (e^{y}) = \frac{\partial}{\partial x} (2 x - y)$

Apply Difference Rule.

$\frac{\partial}{\partial x} e^{y} = \frac{\partial}{\partial x} 2 x - \frac{\partial}{\partial x} y$

Apply Constant Multiple Rule.

$\frac{\partial}{\partial x} e^{y} = 2 \frac{\partial}{\partial x} x - \frac{\partial}{\partial x} y$

Apply Power Rule.

$\frac{\partial}{\partial x} e^{y} = 2 - \frac{\partial}{\partial x} y$

Step 2: Differentiate the y-terms.

Original

$\frac{\partial}{\partial x} e^{y} = 2 - \frac{\partial}{\partial x} y$

Implicitly differentiate e^y using the natural exponent rule.

$\frac{\partial}{\partial x} e^{y} = \frac{\partial}{\partial y} e^{y} \frac{\partial y}{\partial x} = e^{y} \frac{\partial y}{\partial x}$

Implicitly differentiate $\frac{\partial}{\partial x} y$ using the power rule.

$\frac{\partial}{\partial x} y = \frac{\partial}{\partial y} y \frac{\partial y}{\partial x} = 1 \frac{\partial y}{\partial x}$

$1 \frac{\partial y}{\partial x}$

Substitute into the original equation:

$e^{y} \frac{\partial y}{\partial x} = 2 - 1 \frac{\partial y}{\partial x}$

Step 3: Solve for $\frac{\partial y}{\partial x}$ and simplify.

Original

$e^{y} \frac{\partial y}{\partial x} = 2 - 1 \frac{\partial y}{\partial x}$

Add $1 \frac{\partial y}{\partial x}$ to both sides:

$e^{y} \frac{\partial y}{\partial x} + 1 \frac{\partial y}{\partial x} = 2$

Factor out $\frac{\partial y}{\partial x}$ from the left-hand side:

$\frac{\partial y}{\partial x} (e^{y} + 1) = 2$

Divide both sides by e^y + 1.

$\frac{\partial y}{\partial x} = \frac{2}{e^{y} + 1}$

Related Links:
Math
algebra
Combination of Functions
Composition of Functions
Calculus Topics

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