Implicit Differentiation - AP Calculus AB Study Guide

The Core Idea: Implicit Differentiation

Some functions are not defined explicitly in the form $y=f(x)$, but are instead defined by a relationship between $x$ and $y$, such as $x^2+y^2=25$. These are called implicitly defined functions. While we may not be able to easily solve for $y$, we can still find its rate of change with respect to $x$, which is the derivative $\frac{dy}{dx}$.

Implicit differentiation is the technique used to find this derivative. The core principle is to treat $y$ as a function of $x$ and apply the chain rule whenever we differentiate a term involving $y$. This process allows us to calculate the slope of the tangent line to a curve at any point, even when the curve cannot be represented by a single, simple function of $x$.

The Key Process: Implicit Differentiation

Implicit differentiation is a procedure rather than a single formula. The goal is to find $\frac{dy}{dx}$ for an implicitly defined function.

The process involves these steps:

1. Differentiate both sides of the equation with respect to $x$.

2. Apply the chain rule to all terms involving $y$. Remember that $y$ is a function of $x$, so its derivative is $\frac{dy}{dx}$. For any function of $y$, say $f(y)$, its derivative with respect to $x$ is $f^{\prime }(y)\cdot \frac{dy}{dx}$.

3. Isolate the $\frac{dy}{dx}$ terms. After differentiating, move all terms containing $\frac{dy}{dx}$ to one side of the equation and all other terms to the opposite side.

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

5. Solve for $\frac{dy}{dx}$ by dividing both sides of the equation by the factor that was multiplied by $\frac{dy}{dx}$. The resulting expression for $\frac{dy}{dx}$ may involve both $x$ and $y$.

Understanding the Role of the Chain Rule

The chain rule is the fundamental basis for implicit differentiation. When we differentiate an expression with respect to $x$, any term involving $y$ must be treated as a composite function because $y$ itself is considered a function of $x$.

Consider the term $y^3$. We are differentiating this with respect to $x$, not $y$.

• The "outer function" is $(\cdot {)}^3$.

• The "inner function" is $y(x)$.

According to the chain rule, the derivative is the derivative of the outer function (evaluated at the inner function) multiplied by the derivative of the inner function.

$$\frac{d}{dx}[y^3]=\underset{\text{Derivative of outer}}{\underbrace{3y^2}}\cdot \underset{\text{Derivative of inner}}{\underbrace{\frac{dy}{dx}}}$$

This principle applies to any function of $y$. For example:

• $\frac{d}{dx}[\sin (y)]=\cos (y)\cdot \frac{dy}{dx}$

• $\frac{d}{dx}[e^y]=e^y\cdot \frac{dy}{dx}$

Forgetting to multiply by $\frac{dy}{dx}$ is the most common error in this process and represents a fundamental misunderstanding of the chain rule's application.

Core Concepts & Rules

• Implicit differentiation is the technique used to find the derivative of a function when $y$ is not explicitly defined in terms of $x$.

• The chain rule is the foundational principle for implicit differentiation.

• When differentiating any term containing $y$ with respect to $x$, you must multiply by $\frac{dy}{dx}$.

• The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

• The final expression for $\frac{dy}{dx}$ will often contain both $x$ and $y$ variables. To find a numerical slope, you need the coordinates of a specific point $(x,y)$.

Step-by-Step Example 1: Finding $dy/dx$

Problem: Find $\frac{dy}{dx}$ for the equation $x^3+y^3=8$.

Step 1: Differentiate both sides with respect to $x$

$$\frac{d}{dx}[x^3+y^3]=\frac{d}{dx}[8]$$

Step 2: Apply differentiation rules to each term

For the $x^3$ term, we use the power rule:

$$\frac{d}{dx}[x^3]=3x^2$$

For the $y^3$ term, we use the power rule combined with the chain rule:

$$\frac{d}{dx}[y^3]=3y^2\cdot \frac{dy}{dx}$$

For the constant $8$, the derivative is $0$.

Putting it all together, the differentiated equation is:

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

Step 3: Isolate the $\frac{dy}{dx}$ term

Subtract $3x^2$ from both sides:

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

Step 4: Solve for $\frac{dy}{dx}$

Divide both sides by $3y^2$:

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

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

Step-by-Step Example 2: Exam-Style Application

Problem: Find the equation of the tangent line to the curve defined by $x^2+xy-y^3=7$ at the point $(3,2)$.

Step 1: Differentiate both sides with respect to $x$

$$\frac{d}{dx}[x^2+xy-y^3]=\frac{d}{dx}[7]$$

Step 2: Apply differentiation rules to each term

• $\frac{d}{dx}[x^2]=2x$

• For the $xy$ term, we must use the product rule: $\frac{d}{dx}[xy]=(\frac{d}{dx}[x])(y)+(x)(\frac{d}{dx}[y])=(1)(y)+(x)(\frac{dy}{dx})=y+x\frac{dy}{dx}$

• For the $-y^3$ term, use the chain rule: $\frac{d}{dx}[-y^3]=-3y^2\frac{dy}{dx}$

• $\frac{d}{dx}[7]=0$

The full differentiated equation is:

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

Step 3: Isolate the $\frac{dy}{dx}$ terms

Move all terms without $\frac{dy}{dx}$ to the right side:

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

Step 4: Factor out $\frac{dy}{dx}$

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

Step 5: Solve for $\frac{dy}{dx}$

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

Step 6: Evaluate the slope at the point $(3,2)$

Substitute $x=3$ and $y=2$ into the expression for $\frac{dy}{dx}$:

$${\frac{dy}{dx}|}_{(3,2)}=\frac{-2(3)-(2)}{(3)-3(2{)}^2}=\frac{-6-2}{3-3(4)}=\frac{-8}{3-12}=\frac{-8}{-9}=\frac{8}{9}$$

The slope of the tangent line at $(3,2)$ is $\frac{8}{9}$.

Step 7: Write the equation of the tangent line

Using the point-slope form $y-y_1=m(x-x_1)$:

$$y-2=\frac{8}{9}(x-3)$$

Using Your Calculator

Implicit differentiation is a purely analytical technique. A graphing calculator like the TI-84 cannot perform the symbolic steps required to find the derivative $\frac{dy}{dx}$ from an implicit equation. You must learn and perform the process by hand.

The calculator's role is limited to arithmetic verification. For instance, in Example 2, after finding the slope $\frac{dy}{dx}=\frac{-2x-y}{x-3y^2}$, you can use the calculator to substitute the values $x=3$ and $y=2$ to compute the final numerical slope, which is especially helpful if the numbers are complex. You would enter `(-23 - 2)/(3 - 32^2)to get the result $8/9. The calculator is a tool for checking your arithmetic, not for performing the calculus.

AP Exam Quick Hit

Common Question Types

• Find $\frac{dy}{dx}$: Given an equation relating $x$ and $y$, find a general expression for $\frac{dy}{dx}$. Example: $Findthederivative$\frac{dy}{dx}for the curve $e^y = x^2 + y.`

• Evaluate the slope at a point: Find the slope of the tangent line to an implicitly defined curve at a given point. Example: What is the slope of the line tangent to the curve $y^2 + 2x = 2y + x^2 at the point $(2,2)$?`

• Find the equation of a tangent line: Find the full equation of the tangent line to a curve at a given point, as shown in Example 2 above.

Common Mistakes

• Forgetting $\frac{dy}{dx}$: The most frequent error is differentiating a $y$ term but forgetting to apply the chain rule by multiplying by $\frac{dy}{dx}$. For example, differentiating $y^2$ as $2y$ instead of the correct $2y\frac{dy}{dx}$.

• Incorrectly Applying the Product Rule: For a term like $x^{2}y$, students often forget to use the product rule, differentiating it as $2x\frac{dy}{dx}$. The correct derivative is $2xy+x^2\frac{dy}{dx}$.

• Algebraic Errors in Isolation: After correctly differentiating, students make errors when rearranging the equation to solve for $\frac{dy}{dx}$, such as sign errors when moving terms across the equals sign or mistakes in factoring.

• Differentiating a Constant to 1: Incorrectly differentiating a constant term (e.g., the $7$ in $x^2+y^2=7$) to $1$ instead of $0$.