Domanda del libro di testo
The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle’s
b. area?
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Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 3, Problema 3.7.61b
Computer Explorations
Use a CAS to perform the following steps in Exercises 55–62.
b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.
2y² + (xy)¹/³ = x² + 2, P(1,1)
Guida verificata passo dopo passo1
Start by differentiating both sides of the equation 2y² + (xy)^(1/3) = x² + 2 with respect to x. Remember that y is a function of x, so you'll need to use implicit differentiation.
Differentiate the left side: For 2y², use the chain rule to get 4y(dy/dx). For (xy)^(1/3), use the chain rule and product rule: differentiate (xy)^(1/3) with respect to x, which involves differentiating the inside function xy.
Differentiate the right side: The derivative of x² is 2x, and the derivative of the constant 2 is 0.
Set the derivatives equal: Combine the results from the left and right sides to form the equation 4y(dy/dx) + (1/3)(xy)^(-2/3)(y + x(dy/dx)) = 2x.
Solve for dy/dx: Isolate dy/dx on one side of the equation. Substitute the point P(1,1) into the equation to evaluate the derivative at that point.
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Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, typically x, while treating other variables as implicit functions of x. This method is essential when dealing with equations where y cannot be easily isolated.
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Percorso guidato
05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This rule is crucial when applying implicit differentiation, especially when differentiating terms like (xy)^(1/3), where both x and y are functions of another variable.
Video consigliato:
05:02Intro to the Chain Rule
Evaluating Derivatives at a Point
Once the derivative dy/dx is found using implicit differentiation, it can be evaluated at a specific point to find the slope of the tangent line at that point. This involves substituting the coordinates of the given point, such as P(1,1), into the derivative formula. This step is important for understanding the behavior of the function at specific locations on its graph.
Video consigliato:
04:50Critical Points
Pratica correlata
Domanda del libro di testo
Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.
x ƒ(x) g(x) ƒ'(x) g'(x)
0 1 1 -3 1/2
1 3 5 1/2 -4
Find the first derivatives of the following combinations at the given value of x.
c. ƒ(x) , x = 1
g(x) + 1
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Domanda del libro di testo
Computer Explorations
Use a CAS to perform the following steps in Exercises 55–62.
b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.
x√(1 + 2y) + y = x², P(1,0)
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Domanda del libro di testo
The folium of Descartes (See Figure 3.27)
c. Find the coordinates of the point A in Figure 3.29 where the folium has a vertical tangent line.
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Domanda del libro di testo
Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure θ is
A = (1/2) ab sinθ.
b. How is dA/dt related to dθ/dt and da/dt if only b is constant?
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Domanda del libro di testo
Differentiability and Continuity on an Interval
Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be
c. neither continuous nor differentiable?
Give reasons for your answers.
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