3.2 Implicit Differentiation

Welcome back to AP Calculus with Fiveable! This topic focuses on implicit differentiation and taking derivatives when variables cannot be isolated. We worked through the chain rule in the last lesson, so let's build on that idea. 🙌

🔎 Implicit Differentiation

We are used to taking the derivative of functions where $y$ is isolated, such as $y = x^2$. This is called an explicit equation!

But what if our equation is not solved for $y$, such as $xy^2 = xy + 1$? This is an implicit equation, so enter implicit differentiation! 🔄 Implicit differentiation allows us to find the derivative of equations where $y$ is not explicitly expressed in terms of $x$.

The general idea is to differentiate each side of the equation with respect to the dependent variable, usually $x$. This means that you need to use the chain rule when differentiating for both $x$ and $y.$ Then, you can isolate $\frac{dy}{dx}$ to get a final answer.

🪜 Implicit Differentiation Steps

Here are a couple of steps to successfully differentiate an equation implicitly:

1. Notate that you’re differentiating. Differentiate both sides of the equation with respect to $x$.
2. Apply your knowledge of derivative rules, such as the power rule and chain rule.
3. Isolate $\frac{dy}{dx}$.

You will often find yourself factoring out $\frac{dy}{dx}$ to isolate it!

🧐 Implicit Differentiation Walkthrough

Implicit Differentiation is best learned through practice, so let's go through and find $\frac{dy}{dx}$ for the unit circle, $x^2 + y^2 = 1$.

First, we need to notate that we are differentiating.

Next, we can apply our knowledge of the Power Rule and the Chain Rule.

Because $\frac{dx}{dx} = 1$, we can leave it off when doing the chain rule for this equation.

Lastly, isolate $\frac{dy}{dx}$ and you get the final answer of…

We can see this holds true in the graph below! The slope of the graph at any point can be represented by $\frac{dy}{dx} = \frac{-x}{y}$.

For example, at the point $\textcolor{green}{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)}$, we can calculate…

At the point $\textcolor{black}{\left(\frac{-3}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)}$, we can calculate…

Graph created with Desmos

🧮 Implicit Differentiation Practice Problems

Let’s work on a few questions and make sure we have the concept down!

The following free-response question (FRQ) is based on from the 2015 AP Calculus AB examination administered by College Board. All credit to College Board.

1) Implicit Derivatives and Tangent Lines

Consider the curve given by $y^3 -xy = 2$.

(a) Calculate $\frac {dy}{dx}$ .

(b) Write an equation for the line tangent to the curve at the point $(-1,1)$

a) Calculate the derivative

Let’s work through calculating the derivative implicitly with the chain rule first.

We have to use the product rule for the second term here!

Now we can finally isolate $\frac{dy}{dx}$.

Therefore,

b) Equation for the tangent line

We have almost all of the information needed to plug into this equation! $y - y_1 = m(x - x_1)$

$x_1 = -1$, $y_1 = 1$, and now we need to solve for $\frac{dy}{dx}|_{(-1,1)}$.

All you have to do is plug in the values for $x$ and $y$ into the derivative equation we just calculated.

Therefore,

Now we can plug in all of the information! Our equation for the tangent line to the curve at the point $(-1,1)$ is:

Way to go! These answers would accumulate 3/3 points on the FRQ. 🤜

🌟 Closing

Great work! 🙌 Implicit differentiation is a powerful tool in your calculus toolkit. 🧰 It allows you to tackle equations that may seem tricky at first glance. You can anticipate encountering questions involving implicit differentiation on the exam, both in multiple-choice and as part of a free response.

Image Courtesy of Giphy