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Location Engineering Building | room 332
Phone 435-797-1494
Contact Christian Bolander
Email christian.bolander@usu.edu

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Implicit Differentiation

Sometimes we are faced with equations that do not explicitly define y (the dependent variable) as a function of x (the independent variable), nor are they easily rearranged to have y in terms of x. Assuming that the dependent variable is differentiable, we can differentiate the function and implicitly find $\frac{d y}{d x}$ and solve for it.

Engineering Context:

In engineering, implicit differentiation can be used to determine relationships between the variables in an implicit equation. Thus, whenever an equation can be written easily in implicit form, that is

$ f(x, y, \ldots) = 0 $

the relationship between any of the variables x, y, etc. can be found. Implicit functions don’t appear all that often in engineering practice, but its important to keep this concept as a tool that can be used when the need arises, particularly when dealing with unusual or complex functions.

The Essentials

When we find ourselves with a function such as $x^{2} + y^{3} = 4$ we will use the process of implicit differentiation to solve. The steps are as follows:

1. Differentiate the entire equation with respect to x. Don’t forget the chain rule! (Technically, the chain rule says that we should include a $\frac{d x}{d x}$ with each x term as well, but those terms trivially reduce to 1 so we won’t bother.)

\frac{d}{d x} (x^{2} + y^{3}) = \frac{d}{d x} (4)

2 x + 3 y^{2} \frac{dy}{dx} = 0

2. Solve for $\frac{d y}{d x}$

(a) Move the terms not associated with $\frac{d y}{d x}$ to the right.

2 x + 3 y^{2} \frac{dy}{dx} = 0

3 y^{2} \frac{dy}{dx} = - 2 x

(b) Divide both sides of the equation by $3 y^{2}$

3 y^{2} \frac{dy}{dx} = - 2 x

\frac{dy}{dx} = - \frac{2 x}{3 y^{2}}

As you can see we have found the derivative of y with respect to x. Note that the derivative is a function of both x and y. This is the process of implicit differentiation.

A Deeper Dive

We are generally most comfortable working with explicit functions, such as y = mx + b. This is called an explicit function or equation because y is defined directly as a function of x. Sometimes it may be difficult to identify when the method of implicit differentiation should be used, so below is a table to help begin to identify explicit vs. implicit functions so you can know when to use implicit differentiation.

Explicit Function Examples	Implicit Function Examples
$y = 2 x + 7$	$x^{2} + y^{2} = 25$
$y = x^{3} + 4 x - 12$	$4 x^{4} - 3 y^{2} + 9 x = 23 - 12 y$
$x = a \sin θ$	$v^{2} w + v w^{2} = v w - 4$

Practice

Use implicit differentiation to find $\frac{d y}{d x}$ .

x^{2} + x y + y^{2} = 17

Solution:

First, differentiate the equation with respect to x. We must use the product rule to differentiate xy. Remember, this will include a $\frac{d x}{d x}$ with the x term, but it simplifies to 1.

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

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

Second, solve for $\frac{d y}{d x}$ .

x \frac{dy}{dx} + 2 y \frac{dy}{dx} = - 2 x - y

(2 y + x) \frac{dy}{dx} = - 2 x - y