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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 πŸš€

Hey there, future calculus masters! πŸ‘‹ 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? πŸ€”

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. πŸ”„

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(otherΒ side)\frac{d}{dx} (\text{equation}) = \frac{d}{dx} (\text{other side})dxd​(equation)=dxd​(otherΒ 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 yβ€²y'yβ€².
  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 β­•

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

  1. Notate:
latex
$$
\frac{d}{dx}(x^2 + y^2) = \f...
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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​. πŸš€

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

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

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

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