Domanda del libro di testo
60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
a. Find equations of all lines tangent to the curve at the given value of x.
4x³ =y²(4−x); x=2 (cissoid of Diocles)
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 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)
Guida verificata passo dopo passo1
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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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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Percorso guidato
05:14
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
Pratica correlata
Domanda del libro di testo
79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)
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Domanda del libro di testo
Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.
a. Show that the stones reach their high points at the same time.
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Domanda del libro di testo
Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.
f(x) = -7x; P(-1,7)
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Domanda del libro di testo
Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.
73. {Use of Tech} Graph the following functions and determine the location of the vertical tangent lines.
a. f(x) = (x-2)^1/3
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Domanda del libro di testo
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = √3x; a= 12
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