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  1. AP Calculus
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Implicit Differentiation

Benjamin Wright

Benjamin Wright

7 min read

Next Topic - Differentiating Inverse Functions

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Study Guide Overview

This guide covers implicit differentiation, focusing on how to find racdydxrac{dy}{dx}racdydx when y is not explicitly defined in terms of x. It outlines the steps for implicit differentiation, including using the chain rule and product rule. The guide provides examples, practice problems involving tangent lines, and common mistakes to avoid. Finally, it offers exam tips and strategies, highlighting high-priority topics like combining implicit differentiation with related rates problems.

AP Calculus AB/BC: Implicit Differentiation - Your Ultimate Guide 🚀

Hey there, future calculus masters! 👋 Let's dive into implicit differentiation, a key technique that'll help you ace those tricky problems. This guide is designed to be your go-to resource, especially the night before the exam. Let's make sure you're feeling confident and ready!

Implicit Differentiation: Unlocking Hidden Derivatives

What is Implicit Differentiation? 🤔

We're used to explicit equations like y=x2y = x^2y=x2, where yyy is isolated. But what about equations like xy2=xy+1xy^2 = xy + 1xy2=xy+1? That's where implicit differentiation comes in! It's a method to find derivatives when yyy isn't explicitly defined in terms of xxx. Think of it as finding the slope of a curve even when the equation is a bit tangled. 🔄

Key Concept

Key Idea: Differentiate both sides of the equation with respect to xxx, remembering to use the chain rule for both xxx and yyy. Then, solve for dydx\frac{dy}{dx}dxdy​.

Memory Aid

Chain Rule Reminder: When differentiating a term involving yyy, remember to multiply by dydx\frac{dy}{dx}dxdy​ because yyy is a function of xxx!

Exam Tip

Steps for Implicit Differentiation

  1. Notate: Indicate that you're differentiating both sides with respect to xxx. ddx(equation)=ddx(other side)\frac{d}{dx} (\text{equation}) = \frac{d}{dx} (\text{other side})dxd​(equation)=dxd​(other side)
  2. Differentiate: Apply derivative rules (power rule, product rule, chain rule). Remember, dxdx=1\frac{dx}{dx} = 1dxdx​=1, but dydx\frac{dy}{dx}dxdy​ remains as dydx\frac{dy}{dx}dxdy​ or y′y'y′.
  3. Isolate: Solve for dydx\frac{dy}{dx}dxdy​. You'll often need to factor out dydx\frac{dy}{dx}dxdy​.

Example: The Unit Circle ⭕

Let's find dydx\frac{dy}{dx}dxdy​ for x2+y2=1x^2 + y^2 = 1x2+y2=1:

  1. Notate:
<math-block>\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(1)</math-block>
  1. Differentiate:
<math-block>2x\frac{dx}{dx} + 2y\frac{dy}{dx} = 0</math-block>
<math-block>2x + 2y\frac{dy}{dx} = 0</math-block>
  1. Isolate:
<math-block>2y\frac{dy}{dx} = -2x</math-block>
<math-block>\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}</math-block>
Quick Fact

Quick Check: The derivative dydx=−xy\frac{dy}{dx} = \frac{-x}{y}dxdy​=y−x​ gives the slope of the unit circle at any point (x,y)(x, y)(x,y).

For instance, at the point (12,12)\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)(2​1​,2​1​), dydx=−1\frac{dy}{dx} = -1dxdy​=−1, and at the point (−32,−12)\left(\frac{-3}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)(2​−3​,2​−1​), dydx=−3\frac{dy}{dx} = -\sqrt{3}dxdy​=−3​.

Screen Shot 2023-12-13 at 6.19.58 PM.png

Graph of x2+y2=1x^2 + y^2 = 1x2+y2=1

Graph created with Desmos


✍️ Implicit Differentiation Practice Problems

Let's solidify your understanding with some practice problems. These are designed to mimic what you'll see on the AP exam.

1) Implicit Derivatives and Tangent Lines

Consider the curve given by y3−xy=2y^3 - xy = 2y3−xy=2.

(a) Calculate dydx\frac{dy}{dx}dxdy​.

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

a) Calculate the derivative

  1. Notate:
<math-block>\frac{d}{dx}(y^3 - xy) = \frac{d}{dx}(2)</math-block>
  1. Differentiate:
<math-block>3y^2\frac{dy}{dx} - (x\frac{dy}{dx} + y) = 0</math-block>
  1. Isolate:
<math-block>3y^2\frac{dy}{dx} - x\frac{dy}{dx} - y = 0</math-block>
<math-block>\frac{dy}{dx}(3y^2 - x) = y</math-block>
<math-block>\frac{dy}{dx} = \frac{y}{3y^2 - x}</math-block>

b) Equation for the tangent line

Exam Tip

Tangent Line Equation: Remember the point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​), where mmm is the slope (derivative) at the point (x1,y1)(x_1, y_1)(x1​,y1​).

We have x1=−1x_1 = -1x1​=−1, y1=1y_1 = 1y1​=1. Now find the slope dydx\frac{dy}{dx}dxdy​ at (−1,1)(-1, 1)(−1,1):

<math-block>\frac{dy}{dx} = \frac{1}{3(1)^2 - (-1)} = \frac{1}{4}</math-block>

The tangent line equation is:

<math-block>y - 1 = \frac{1}{4}(x + 1)</math-block>
Common Mistake

Common Mistake: Forgetting to use the product rule when differentiating terms like xyxyxy or forgetting dydx\frac{dy}{dx}dxdy​ when differentiating y. Always double-check your steps!

Memory Aid

Product Rule Reminder: ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}dxd​(uv)=udxdv​+vdxdu​. Don't forget it when you have two functions multiplied together!

You aced it! 3/3 points on the FRQ. 🌟


🎉 Closing Thoughts

You've now got the hang of implicit differentiation! It's a powerful tool that will come up in many different forms on the exam. Keep practicing, and you'll be ready for anything. Remember, you've got this! 💪

https://zupay.blob.core.windows.net/resources/files/0baca4f69800419293b4c75aa2870acd_5cf169_3316.gif?cid=7941fdc64i8t3xwgu78ov7fo1wvrrynfxrd4d9loorgxbu2d&ep=v1_gifs_search&rid=giphy.gif&ct=g

Encouraging GIF with animated ice cream

Image Courtesy of Giphy


Final Exam Focus 🎯

  • High-Priority Topics: Implicit differentiation is frequently combined with related rates and tangent line problems.

  • Common Question Types: Expect to find dydx\frac{dy}{dx}dxdy​, use it to find tangent lines, and solve related rates problems.

  • Time Management: Practice solving problems quickly. Focus on identifying the key steps and avoid getting bogged down in algebra.

  • Common Pitfalls: Watch out for product rule and chain rule errors. Always double-check your work.

Exam Tip

Exam Strategy: If you get stuck, write down what you know and try to apply the steps of implicit differentiation. Partial credit is often given for correct setup and differentiation.


Practice Question

Practice Questions

Multiple Choice Questions

  1. If x2+y2=25x^2 + y^2 = 25x2+y2=25, then dydx\frac{dy}{dx}dxdy​ at the point (3, 4) is: (A) -3/4 (B) 3/4 (C) -4/3 (D) 4/3

  2. If xy+y2=3xy + y^2 = 3xy+y2=3, then dydx\frac{dy}{dx}dxdy​ at the point (2, 1) is: (A) -1/4 (B) -1/3 (C) -1/2 (D) 1/2

Free Response Question

Consider the curve defined by x3+y3=6xyx^3 + y^3 = 6xyx3+y3=6xy.

(a) Find dydx\frac{dy}{dx}dxdy​.

(b) Write an equation for the line tangent to the curve at the point (3, 3).

(c) Find all points on the curve where the tangent line is horizontal.

FRQ Scoring Breakdown

(a) Finding the derivative (5 points): * 1 point: Correctly applying the power rule and chain rule on x3x^3x3 and y3y^3y3 * 2 points: Correctly applying the product rule on 6xy * 1 point: Correctly isolating dydx\frac{dy}{dx}dxdy​ * 1 point: Correctly simplifying the final answer

(b) Tangent Line (2 points): * 1 point: Correctly evaluating the derivative at (3,3) * 1 point: Correctly writing the equation of the tangent line

(c) Horizontal Tangents (2 points): * 1 point: Setting dydx\frac{dy}{dx}dxdy​ equal to 0 * 1 point: Correctly finding all points

Answers

MCQ

  1. (A)
  2. (B)

FRQ

(a) dydx=2y−x2y2−2x\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}dxdy​=y2−2x2y−x2​

(b) y−3=−1(x−3)y - 3 = -1(x - 3)y−3=−1(x−3)

(c) (0,0) and (4, 2)

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Previous Topic - The Chain RuleNext Topic - Differentiating Inverse Functions

Question 1 of 7

Given the equation x2+y2=25x^2 + y^2 = 25x2+y2=25, find dydx\frac{dy}{dx}dxdy​. 🚀

dydx=xy\frac{dy}{dx} = \frac{x}{y}dxdy​=yx​

dydx=−xy\frac{dy}{dx} = -\frac{x}{y}dxdy​=−yx​

dydx=yx\frac{dy}{dx} = \frac{y}{x}dxdy​=xy​

dydx=−yx\frac{dy}{dx} = -\frac{y}{x}dxdy​=−xy​