# What is Implicit Differentiation?

### Quick Overview

• Implicit differentiation allows us to find derivatives of implicit curves.
• Implicit differentiation relies on the chain rule.

### Implicit and Explicit Functions

Explicit Functions: When a function is written so that the dependent variable is isolated on one side of the equation, we call it an explicit function.

That is, an explicit function is written $$y = f(x)$$. All of the functions that we've worked with so far have been explicit.

Implicit Functions: When the dependent variable is not isolated, we refer to the equation as being implicitly defined and the $$y$$-variable as an implicit function.

### Circles, Ellipses and Hyperbolas

You are probably most familiar with implicit functions that describe the conic sections: circles, ellipses and hyperbolas.

The graphs of each of these is shown below. Notice how each equation can't be solved explicitly for $$y$$.

### Differentiating Implicitly---Basic Idea

If $$y$$ is a function, then $$f(y)$$ is a composite function. That means finding the derivative of $$f(y)$$ requires the chain rule, like this:

$$\frac d {dx}\big(f(y)\big) = f'(y)\cdot \frac{dy}{dx}.$$

### Examples

##### Example 1

Suppose $$f(y) = y^3$$. Evaluate $$\displaystyle \frac d {dx}\left(y^3\right)$$..

$$\frac d {dx}\left(y^3\right) = 3y^2\cdot \frac{dy}{dx}$$

##### Example 2

Evaluate $$\displaystyle \frac d {dx}\left(\sin y\right)$$.

$$\frac d {dx}\left(\sin y\right) = (\cos y)\,\frac{dy}{dx}$$

This use of the chain rule is the basic idea behind implicit differentiation. Usually, we work with implicit equations. We just have to remember to use the chain rule when differentiating something involving $$y$$.

##### Example 3

Find $$\frac{dy}{dx}$$ for the circle $$x^2 + y^2 = 1$$.

Step 1

Differentiate implicitly. Remember, this means $$y$$ is a function, so its derivative is $$\frac{dy}{dx}$$.

\begin{align*} \frac d {dx}(x^2) + \frac d {dx}(y^2) & = \frac d {dx}(1)\\[6pt] 2x + 2y\cdot \frac{dy}{dx} & = 0 \end{align*}

Step 2

Solve the equation for $$\frac{dy}{dx}$$.

\begin{align*} 2x + 2y\cdot \frac{dy}{dx} & = 0\\[6pt] 2y\cdot\frac{dy}{dx} & = - 2x\\[6pt] \frac{dy}{dx} & = -\frac{2x}{2y}\\[6pt] \frac{dy}{dx} & = - \frac x y \end{align*}

$$\displaystyle \frac{dy}{dx} = -\frac x y$$

##### Example 4

The graph of $$8x^3e^{y^2} = 3$$ is shown below. Find $$\displaystyle \frac{dy}{dx}$$.

Step 1

Notice that the left-hand side is a product, so we will need to use the the product rule. Identify the factors that make up the left-hand side.

$$\blue{8x^3}\cdot \red{e^{y^2}} = 3$$

Step 2

Differentiate using the the product rule and implicit differentiation.

\begin{align*}% \blue{24x^2}\cdot e^{y^2} + 8x^3\cdot \red{e^{y^2}\cdot \frac d {dx}\left(y^2\right)} & = 0\\[6pt] 24x^2e^{y^2} + 8x^3e^{y^2}\cdot \red{2y\,\frac{dy}{dx}} & = 0\\[6pt] 24x^2e^{y^2} + 16x^3y\,e^{y^2}\,\frac{dy}{dx} & = 0 \end{align*}

Step 3

Solve the equation for $$\frac{dy}{dx}$$.

\begin{align*} 24x^2e^{y^2} + 16x^3y\,e^{y^2}\,\frac{dy}{dx} & = 0\\[6pt] 16x^3y\,e^{y^2}\,\frac{dy}{dx} & = -24x^2e^{y^2}\\[6pt] \frac{dy}{dx} & = -\frac{24x^2e^{y^2}}{16x^3y\,e^{y^2}} \end{align*}

Step 4

Simplify.

\begin{align*} \frac{dy}{dx} & = -\frac{24x^2e^{y^2}}{16x^3y\,e^{y^2}}\\[6pt] & = -\frac 3 {2xy} \end{align*}

$$\displaystyle \frac{dy}{dx} = -\frac 3 {2xy}$$

##### Example 5

The graph of $$\sin(x + 2y) = \cos x$$ is shown below. Find $$\displaystyle \frac{dy}{dx}$$.

Step 1

Differentiate using implicit differentiation.

$$\cos(x + 2y)\cdot\left(1 + 2\,\frac{dy}{dx}\right) = -\sin x$$

Step 2

Distribute the cosine.

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

Step 3

Solve the equation for $$\frac{dy}{dx}$$.

\begin{align*} \cos(x + 2y) + 2\cos(x + 2y)\frac{dy}{dx} & = -\sin x\\[6pt] 2\cos(x + 2y)\frac{dy}{dx} & = -\sin x - \cos(x + 2y)\\[6pt] \frac{dy}{dx} & = \frac{-\sin x - \cos(x + 2y)}{2\cos(x + 2y)}\\[6pt] \frac{dy}{dx} & = -\frac{\sin x + \cos(x + 2y)}{2\cos(x + 2y)} \end{align*}

$$\displaystyle \frac{dy}{dx} = -\frac{\sin x + \cos(x + 2y)}{2\cos(x + 2y)}$$

##### Example 6

The graph of $$7x^2 + 5x + y^3 - 3y^2 = 12$$ is shown below. Find $$\frac{dy}{dx}$$.

Step 1

Differentiate implicitly.

$$14x + 5 + 3y^2\,\frac{dy}{dx} - 6y\,\frac{dy}{dx} = 0$$

Step 2

Isolate the terms with $$\frac{dy}{dx}$$

$$3y^2\,\frac{dy}{dx} - 6y\,\frac{dy}{dx} = -14x - 5$$

Step 3

Factor out the $$\frac{dy}{dx}$$.

$$\frac{dy}{dx}\left(3y^2 - 6y\right) = -14x - 5$$

Step 4

Solve for $$\frac{dy}{dx}$$.

$$\frac{dy}{dx} = \frac{-14x - 5}{3y^2 - 6y}$$

Step 5

Simplify.

\begin{align*} \frac{dy}{dx} & = \frac{-14x - 5}{3y^2 - 6y}\\[6pt] & = \frac{-(14x + 5)}{-(-3y^2 + 6y)}\\[6pt] & = \frac{14x + 5}{-3y^2 + 6y}\\[6pt] & = \frac{14x + 5}{6y - 3y^2} \end{align*}

$$\displaystyle \frac{dy}{dx} = \frac{14x + 5}{6y - 3y^2}$$

##### Example 7

Find $$\frac{d^2y}{dx^2}$$ for the equation shown below.

$$x^2 + y^2 = 16$$

Step 1

Find the first derivative $$\frac{dy}{dx}$$.

In Example 1 we determined that $$\frac{dy}{dx} = -\frac x y$$.

Step 2

Differentiate again. Notice that the right-hand side will require the quotient rule.

\begin{align*} \frac{d^2y}{dx^2} & = -\frac{y(1) - x\cdot\frac{dy}{dx}}{y^2}\\[6pt] & = -\frac{y - x\cdot\frac{dy}{dx}}{y^2} \end{align*}

Step 3

Replace $$\frac{dy}{dx}$$ with what it is equal to.

\begin{align*} \frac{d^2y}{dx^2} & = -\frac{y - x\cdot\blue{\frac{dy}{dx}}}{y^2}\\[6pt] & = -\frac{y - x\cdot\blue{\left(-\frac x y\right)}}{y^2}\\[6pt] & = -\frac{y + \frac{x^2} y}{y^2}\\[6pt] & = -\frac{\left(y + \frac{x^2} y\right)}{y^2}\cdot \red{\frac y y}\\[6pt] & = -\frac{y^2 + x^2}{y^2} \end{align*}

Step 4

Since $$x^2 + y^2 = 16$$, we can replace the numerator.

\begin{align*} \frac{d^2y}{dx^2} & = -\frac{\blue{x^2 + y^2}}{y^2}\\[6pt] & = -\frac{\blue{16}}{y^2} \end{align*}

$$\displaystyle \frac{d^2y}{dx^2} = -\frac{16}{y^2}$$