Textbook Question
Assume that y = 5x and dx/dt = 2. Find dy/dt
222
views
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- 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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.49
In Exercises 43–50, find by implicit differentiation.
y² = x .
x + 1
Verified step by step guidance1
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.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
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 equations that are difficult or impossible to rearrange.
Recommended video:
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.
Recommended video:
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.
Recommended video:
06:30Derivatives of Other Trig Functions
Related Practice
Textbook Question
In Exercises 41–58, find dy/dt.
y = ((3t − 4) / (5t + 2))⁻⁵
241
views
Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = sin(5√x)
189
views
Textbook Question
Second Derivatives
In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.
2√y = x – y
219
views
Textbook Question
In Exercises 41–58, find dy/dt.
y = (1 + cos(2t))⁻⁴
257
views
Textbook Question
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.
180
views