Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.9.20

Chapter 3, Problem 3.9.20

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = (2√x)/(3(1 + √x))

Verified step by step guidance

Step 1: Begin by identifying the function y = \(\frac{2\sqrt{x}\)}{3(1 + \(\sqrt{x}\))}. This is a rational function where both the numerator and the denominator involve square roots.

Step 2: Apply the quotient rule for derivatives, which states that if you have a function y = \(\frac{u(x)}{v(x)}\), then the derivative dy/dx is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Here, u(x) = 2\(\sqrt{x}\) and v(x) = 3(1 + \(\sqrt{x}\)).

Step 3: Find the derivative of the numerator u(x) = 2\(\sqrt{x}\). The derivative u'(x) can be found using the chain rule. Recall that \(\sqrt{x}\) = x^{1/2}, so u'(x) = 2 \(\cdot\) \(\frac{1}{2}\)x^{-1/2} = x^{-1/2}.

Step 4: Find the derivative of the denominator v(x) = 3(1 + \(\sqrt{x}\)). The derivative v'(x) involves differentiating 1 + \(\sqrt{x}\). The derivative of \(\sqrt{x}\) is \(\frac{1}{2}\)x^{-1/2}, so v'(x) = 3 \(\cdot\) \(\frac{1}{2}\)x^{-1/2} = \(\frac{3}{2}\)x^{-1/2}.

Step 5: Substitute u(x), u'(x), v(x), and v'(x) into the quotient rule formula: dy/dx = \(\frac{x^{-1/2}\) \(\cdot\) 3(1 + \(\sqrt{x}\)) - 2\(\sqrt{x}\) \(\cdot\) \(\frac{3}{2}\)x^{-1/2}}{(3(1 + \(\sqrt{x}\)))^2}. Simplify the expression to find the derivative dy/dx.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that allows us to determine how a function behaves at any given point. The derivative can be interpreted as the slope of the tangent line to the curve of the function at a specific point.

Differential Form

Differential form refers to the expression of derivatives in terms of differentials, typically denoted as dy and dx. In this context, dy represents the change in the function y as x changes, and is calculated using the derivative. This form is particularly useful for understanding how small changes in x affect y.

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. If y = u/v, where u and v are functions of x, the derivative is given by dy/dx = (v(du/dx) - u(dv/dx)) / v². This rule is essential for differentiating functions like y = (2√x)/(3(1 + √x)), where both the numerator and denominator are functions of x.

Recommended video:

06:43

The Quotient Rule

Related Practice

Textbook Question

In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?

54. x³ + y² = sin²y

237

views

Textbook Question

Slopes and Tangent Lines

In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points P₁ and P₂.

226

views

Textbook Question

If L = √(x² + y²), dx/dt = –1, and dy/dt = 3, find dL/dt when x = 5 and y = 12.

193

views

Textbook Question

Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form

P = (nRT / (V − nb)) − (an² / V²),

in which a, b, n, and R are constants. Find dP/dV. (See accompanying figure.)

<IMAGE>

184

views

Textbook Question

Finding Linearizations

In Exercises 1–5, find the linearization L(x) of f(x) at x = a.

f(x) = ∛x, a = −8

204

views

Textbook Question