Textbook Question
Vertical tangent lines
a. Determine the points where the curve x+y³−y=1 has a vertical tangent line (see Exercise 60).
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.8.23a
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
³√x+³√y⁴ = 2;(1,1)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). Remember that \( y \) is a function of \( x \), so when differentiating terms involving \( y \), use implicit differentiation.
Differentiate \( \sqrt[3]{x} \) with respect to \( x \). This is \( \frac{d}{dx}(x^{1/3}) \), which results in \( \frac{1}{3}x^{-2/3} \).
Differentiate \( \sqrt[3]{y^4} \) with respect to \( x \). This requires the chain rule: \( \frac{d}{dx}(y^4)^{1/3} = \frac{1}{3}(y^4)^{-2/3} \cdot 4y^3 \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side. Since the derivative of a constant (2) is 0, you have: \( \frac{1}{3}x^{-2/3} + \frac{4}{3}y^{3}(y^4)^{-2/3} \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. This involves algebraic manipulation to express \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. Instead of solving for y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method allows us to find the derivative dy/dx without isolating y.
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Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When applying implicit differentiation, the chain rule is essential for differentiating terms involving y, as we must multiply by dy/dx when differentiating y with respect to x. This ensures that we account for the relationship between x and y in the differentiation process.
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05:02Intro to the Chain Rule
Evaluating Derivatives at a Point
After finding the derivative dy/dx using implicit differentiation, we often need to evaluate it at a specific point, such as (1,1) in this case. This involves substituting the x and y values into the derived expression for dy/dx to find the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific coordinates.
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04:50Critical Points
Related Practice
Textbook Question
Find an equation of the line tangent to the given curve at a.
y = (x + 5) / (x - 1); a = 3
Textbook Question
Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>
a. Calculate the average velocity of the airliner during the first 1.5 hours of the trip (0 ≤ t ≤ 1.5).
Textbook Question
62–65. {Use of Tech} Graphing f and f'
a. Graph f with a graphing utility.
f(x)=e^−x tan^−1 x on [0,∞)
Textbook Question
If two opposite sides of a rectangle increase in length, how must the other two opposite sides change if the area of the rectangle is to remain constant?
Textbook Question
62–65. {Use of Tech} Graphing f and f'
a. Graph f with a graphing utility.
f(x) = (x−1) sin^−1 x on [−1,1]