Textbook Question
60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
b. Graph the tangent lines on the given graph.
x+y³−y=1; x=1
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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 3.8.23b
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
³√x+³√y⁴ = 2;(1,1)
Verified step by step guidance1
Start by understanding the given equation: \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). This is an implicit function of \( x \) and \( y \).
Differentiate both sides of the equation with respect to \( x \). Remember to use the chain rule for \( \sqrt[3]{y^4} \), which involves differentiating \( y \) with respect to \( x \) (i.e., \( \frac{dy}{dx} \)).
The derivative of \( \sqrt[3]{x} \) with respect to \( x \) is \( \frac{1}{3}x^{-2/3} \). For \( \sqrt[3]{y^4} \), apply the chain rule: \( \frac{4}{3}y^{3} \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side, which is zero, since the derivative of a constant (2) is zero.
Substitute \( x = 1 \) and \( y = 1 \) into the differentiated equation to solve for \( \frac{dy}{dx} \), which represents the slope of the curve at the point (1,1).
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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 one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for curves defined by equations that cannot be easily rearranged.
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05:14
Finding The Implicit Derivative
Slope of a Curve
The slope of a curve at a given point represents the rate of change of the dependent variable with respect to the independent variable at that point. Mathematically, it is found by evaluating the derivative of the function at the specified coordinates. In the context of implicit differentiation, the slope can be determined by substituting the coordinates of the point into the derivative obtained from the implicit differentiation process.
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11:41Summary of Curve Sketching
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that if a function y is defined as a function of u, which in turn is a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This rule is essential in implicit differentiation, where we often encounter nested functions.
Recommended video:
05:02Intro to the Chain Rule
Related Practice
Textbook Question
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
(x+y)^2/3=y; (4, 4)
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Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. d/dx(tan^−1 x) =sec² x
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Textbook Question
Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = √3x; a= 12
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Textbook Question
A differential equation is an equation involving an unknown function and its derivatives. Consider the differential equation y′′(t)+y(t) = 0.
b. Show that y = B cos t satisfies the equation for any constant B.
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Textbook Question
{Use of Tech} A mixing tank A 500-liter (L) tank is filled with pure water. At time t=0, a salt solution begins flowing into the tank at a rate of 5 L/min. At the same time, the (fully mixed) solution flows out of the tank at a rate of 5.5 L/min. The mass of salt in grams in the tank at any time t≥0 is given by M(t) = 250(1000−t)(1−10−³⁰(1000−t)¹⁰) and the volume of solution in the tank is given by V(t) = 500-0.5t.
b. Graph the volume function and verify that the tank is empty when t=1000 min.
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