Domanda del libro di testo
Assume that y = 5x and dx/dt = 2. Find dy/dt
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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 3.49
In Exercises 43–50, find by implicit differentiation.
y² = x .
x + 1
Guida verificata passo dopo passo1
Start by differentiating both sides of the equation with respect to x. The equation is y² = x(x + 1).
Apply the product rule to the right side of the equation. The product rule states that if you have a product of two functions u(x) and v(x), then the derivative is u'(x)v(x) + u(x)v'(x). Here, u(x) = x and v(x) = (x + 1).
Differentiate the left side of the equation with respect to x. Since y is a function of x, use the chain rule: the derivative of y² with respect to x is 2y(dy/dx).
After differentiating both sides, you will have an equation involving dy/dx. Solve this equation for dy/dx to find the derivative of y with respect to x.
Simplify the expression for dy/dx, if possible, to get the final form of the derivative.
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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 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 equations that are difficult or impossible to rearrange.
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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. It states that if a function y is composed of another function 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 is essential in implicit differentiation when dealing with terms involving y.
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05:02Intro to the Chain Rule
Derivative of a Function
The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In the context of implicit differentiation, understanding how to compute derivatives is crucial for finding the slope of the tangent line to the curve defined by the implicit equation.
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06:30Derivatives of Other Trig Functions
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In Exercises 41–58, find dy/dt.
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2√y = x – y
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In Exercises 41–58, find dy/dt.
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a. Graph the function
ƒ(x) = { x, -1 ≤ x < 0
{ tan x, 0 ≤ x ≤ π/4.
b. Is ƒ continuous at x = 0?
c. Is ƒ differentiable at x = 0?
Give reasons for your answers.
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