Implicit Differentiation - AP Calculus BC Study Guide

The Core Idea: Implicit Differentiation

Some functions are defined explicitly, where one variable is isolated and written in terms of another, such as $y=x^2+\sin (x)$. However, many important relationships in mathematics are defined implicitly by an equation relating two variables, such as $x^2+y^2=25$. In such cases, it may be difficult or even impossible to solve for $y$ in terms of $x$. Implicit differentiation is the process that allows us to find the derivative, $\frac{dy}{dx}$, without first solving the equation for $y$.

The fundamental principle that makes this process work is the chain rule. We treat $y$ as a function of $x$ (even if we don't know what that function is) and apply the chain rule whenever we differentiate a term involving $y$. The process involves differentiating both sides of the implicit equation with respect to $x$. The result is a new equation that includes $\frac{dy}{dx}$ as a variable, which we can then solve for algebraically. The final expression for $\frac{dy}{dx}$ will typically be in terms of both $x$ and $y$.

Key Rules: The Process of Implicit Differentiation

Implicit differentiation is not a single formula but a multi-step process based on the chain rule. To find $\frac{dy}{dx}$ for an equation in $x$ and $y$:

1. Differentiate Both Sides: Differentiate both sides of the entire equation with respect to $x$. Remember that the derivative of a sum is the sum of the derivatives, and the derivative of a constant is zero.

   $$\frac{d}{dx}[\text{Left Side}]=\frac{d}{dx}[\text{Right Side}]$$

2. Apply the Chain Rule for $y$ Terms: When differentiating a term involving $y$, treat $y$ as an inner function of $x$. Apply the appropriate differentiation rule (power, trig, exponential, etc.) to the outer function and then multiply by the derivative of the inside, which is $\frac{dy}{dx}$.

   • $\frac{d}{dx}[y^n]=n{y}^{n-1}\cdot \frac{dy}{dx}$

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

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

   For terms involving both $x$ and $y$ (e.g., $x{y}^2$), the product rule or quotient rule must be used in conjunction with this chain rule application.

3. Solve for $\frac{dy}{dx}$: After differentiating, the resulting equation will contain the term $\frac{dy}{dx}$. Use algebraic manipulation to solve for $\frac{dy}{dx}$. This typically involves:

   • Gathering all terms containing $\frac{dy}{dx}$ on one side of the equation.

   • Moving all other terms to the opposite side.

   • Factoring out $\frac{dy}{dx}$.

   • Dividing to isolate $\frac{dy}{dx}$.

Understanding the Role of the Chain Rule

The chain rule is the theoretical foundation for implicit differentiation. The Essential Knowledge statement CHA-3.D.1 specifies that "The chain rule is the basis for implicit differentiation." To understand this, we must always assume that $y$ is some differentiable function of $x$, which we could write as $y(x)$.

Consider the term $y^3$. When we are asked to find $\frac{d}{dx}[y^3]$, we are really being asked to find $\frac{d}{dx}[(y(x){)}^3]$. This is a classic chain rule scenario:

• The outer function is $u^3$. Its derivative is $3u^2$.

• The inner function is $u=y(x)$. Its derivative is $\frac{dy}{dx}$.

Applying the chain rule, $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$, we get:

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

This logic applies to any function of $y$. For a term like $\cos (y)$, we treat it as $\cos (y(x))$.

• The outer function is $\cos (u)$. Its derivative is $-\sin (u)$.

• The inner function is $u=y(x)$. Its derivative is $\frac{dy}{dx}$.

Therefore, by the chain rule:

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

This principle must be applied consistently to every term containing $y$ during the differentiation process.

Core Concepts & Rules

• Purpose: Implicit differentiation is a technique used to find the derivative $\frac{dy}{dx}$ for functions defined by an equation in two variables where $y$ is not explicitly isolated.

• Core Process: The method involves differentiating both sides of the equation with respect to one variable (typically $x$).

• The Chain Rule is Essential: Every time a term involving $y$ is differentiated with respect to $x$, the chain rule must be applied. This results in multiplying the derivative of the term by $\frac{dy}{dx}$.

• Combining Rules: Standard differentiation rules, such as the product rule and quotient rule, are often required in combination with the implicit differentiation process for terms like $x^{2}y$ or $\frac{x}{y}$.

• Final Expression: The resulting expression for $\frac{dy}{dx}$ is typically a function of both $x$ and $y$. This means to evaluate the slope of the tangent line at a point, you need both the $x$- and $y$-coordinates of that point.

Step-by-Step Example 1: Finding the Derivative

Problem: For the curve defined by the equation $x^3+y^3=8$, find an expression for $\frac{dy}{dx}$.

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

Take the derivative of each term on both sides of the equation with respect to $x$.

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

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

Step 2: Apply differentiation rules to each term

• For the $x^3$ term, use the power rule: $\frac{d}{dx}[x^3]=3x^2$.

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

• For the constant term $8$, the derivative is zero: $\frac{d}{dx}[8]=0$.

Step 3: Construct the new equation

Substitute the derivatives back into the equation:

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

Step 4: Algebraically isolate $\frac{dy}{dx}$

First, move the term without $\frac{dy}{dx}$ to the other side.

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

Next, divide by $3y^2$ to solve for $\frac{dy}{dx}$.

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

Step 5: Simplify the final expression

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

This is the derivative of $y$ with respect to $x$ for the given curve.

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

Problem: Find the slope of the tangent line to the curve defined by $y^2+2x=2xy$ at the point $(1,2)$.

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

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

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

Step 2: Apply differentiation rules to each term

• Left side, first term: $\frac{d}{dx}[y^2]=2y\cdot \frac{dy}{dx}$ (Power rule and Chain rule).

• Left side, second term: $\frac{d}{dx}[2x]=2$.

• Right side: The term $2xy$ is a product of $2x$ and $y$. We must use the product rule: $\frac{d}{dx}[uv]=u^{\prime }v+u{v}^{\prime }$.

  • Let $u=2x$ and $v=y$.

  • Then $u^{\prime }=2$ and $v^{\prime }=\frac{dy}{dx}$.

  • $\frac{d}{dx}[2xy]=(2)(y)+(2x)(\frac{dy}{dx})=2y+2x\frac{dy}{dx}$.

Step 3: Construct the new equation

Substitute the derivatives back into the equation:

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

Step 4: Algebraically isolate $\frac{dy}{dx}$

Gather all terms with $\frac{dy}{dx}$ on the left side and all other terms on the right side.

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

Factor out $\frac{dy}{dx}$ from the left side.

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

Divide to solve for $\frac{dy}{dx}$.

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

This can be simplified by factoring out a 2 from the numerator and denominator:

$$\frac{dy}{dx}=\frac{2(y-1)}{2(y-x)}=\frac{y-1}{y-x}$$

Step 5: Evaluate the derivative at the given point $(1,2)$

The slope of the tangent line is the value of the derivative at the specified point. Substitute $x=1$ and $y=2$ into the expression for $\frac{dy}{dx}$.

$$\frac{dy}{dx}{|}_{(1,2)}=\frac{2-1}{2-1}=\frac{1}{1}=1$$

The slope of the tangent line to the curve at the point $(1,2)$ is 1.

Using Your Calculator

Implicit differentiation is a purely analytical technique; there is no built-in calculator function to perform the symbolic differentiation process for you. The calculator's primary role for this topic is to help verify a numerical answer you have already found by hand.

For instance, to verify the result from Example 2 ($slope=1$ at $(1,2)$ for $y^2+2x=2xy$):

1. Graph the Relation: Some modern graphing calculators have features to graph implicit equations directly. If so, you can enter the equation $y^2+2x=2xy$.

2. Use a Tangent Line Feature: Access the calculator's DRAW menu and select the $Tangent($ function.

3. Specify the Point: Move the cursor as close as possible to the point $(1,2)$ on the graphed curve and press ENTER.

4. Check the Output: The calculator will draw the tangent line and display its equation, often in the form $y=mx+b$. The value of $m$ should be very close to the $1$ that you calculated analytically.

This graphical check provides confidence in your analytical work but does not replace the need to master the step-by-step process of implicit differentiation.

AP Exam Quick Hit

Common Question Types

• Find a general expression for $\frac{dy}{dx}$: Given an implicit relation, find the derivative in terms of $x$ and $y$.

  • Example: "For the curve given by $e^y=x^2+y$, find $\frac{dy}{dx}$."

• Find the slope of the tangent line at a point: Given an implicit relation and a point $(a,b)$ on the curve, calculate the numerical value of $\frac{dy}{dx}$ at that point.

  • Example: "What is the slope of the line tangent to the curve $3y^2-2x^2=6-2xy$ at the point $(3,2)$?"

• Find the second derivative $\frac{d^{2}y}{d{x}^2}$: After finding $\frac{dy}{dx}$, differentiate the expression again with respect to $x$, which will require another round of implicit differentiation and substitution.

  • Example: "Given $x^2+y^2=9$, find $\frac{d^{2}y}{d{x}^2}$ in terms of $x$ and $y$."

Common Mistakes

• Forgetting the $\frac{dy}{dx}$ Factor: This is the most frequent error. Students correctly differentiate a term like $y^2$ to $2y$ but forget to multiply by $\frac{dy}{dx}$ as required by the chain rule.

• Incorrectly Applying the Product Rule: For terms like $xy$, students may differentiate it as $(1)(\frac{dy}{dx})$, forgetting the full product rule which is $(1)(y)+(x)(\frac{dy}{dx})$.

• Algebraic Errors in Isolation: After correctly differentiating, students often make mistakes when rearranging the equation to solve for $\frac{dy}{dx}$, such as errors with distributing negative signs or factoring.

• Treating $y$ as a Constant: Some students mistakenly treat $y$ as a constant when differentiating with respect to $x$, leading them to an incorrect derivative of 0 for any term containing $y$.

• Stopping Before Evaluating: In problems that ask for the slope at a specific point, students may correctly find the general expression for $\frac{dy}{dx}$ but forget to substitute the given $x$ and $y$ values to find the final numerical answer.