Textbook Question
Using the Alternative Formula for Derivatives
Use the formula
f'(x) = lim (z → x) (f(z) − f(x)) / (z − x)
to find the derivative of the functions in Exercises 23–26.
g(x) = 1 + √x
239
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 3.9.21
Derivatives in Differential Form
In Exercises 17–28, find dy.
2y³/² + xy − x = 0
Verified step by step guidance1
First, identify the given equation: 2y^(3/2) + xy - x = 0. We need to find dy, which involves differentiating the equation with respect to x.
Apply implicit differentiation to each term of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, use the chain rule.
Differentiate the first term, 2y^(3/2), with respect to x. Using the chain rule, this becomes (3/2) * 2 * y^(1/2) * dy/dx.
Differentiate the second term, xy, with respect to x. This requires the product rule: differentiate x to get 1, and differentiate y to get dy/dx, resulting in y + x * dy/dx.
Differentiate the third term, -x, with respect to x, which simply becomes -1. Combine all differentiated terms and solve for dy/dx to find dy.
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 find the derivative of a function when it is not explicitly solved for one variable in terms of another. In equations where y is not isolated, we differentiate both sides with respect to x, treating y as a function of x, and apply the chain rule to terms involving y.
Recommended video:
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 differentiating terms like y^(3/2) with respect to x, where y is a function of x.
Recommended video:
05:02Intro to the Chain Rule
Solving for dy
After applying implicit differentiation, the goal is to solve for dy/dx, which represents the derivative of y with respect to x. This involves isolating dy/dx on one side of the equation, often requiring algebraic manipulation to combine like terms and factor out dy/dx, ultimately expressing it in terms of x and y.
Recommended video:
5:02Solving Logarithmic Equations
Related Practice
Textbook Question
The general polynomial of degree n has the form
P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀,
where aₙ ≠ 0. Find P'(x).
203
views
Textbook Question
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
g(x) = 8 / x², (2, 2)
230
views
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
f(x) = √(7 + x sec x)
274
views
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = x² sin² (2x²)
364
views
Textbook Question
Finding Derivative Values
In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.
f(u) = u + 1/cos²u, u = g(x) = πx, x = 1/4
258
views