Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.1.18

Chapter 3, Problem 3.1.18

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.

f(x) = √(x + 1), (8, 3)

Verified step by step guidance

To find the slope of the function's graph at the given point, we need to compute the derivative of the function f(x) = √(x + 1). The derivative, f'(x), represents the slope of the tangent line at any point x.

Use the chain rule to differentiate f(x) = (x + 1)^(1/2). The chain rule states that if you have a composite function g(h(x)), the derivative is g'(h(x)) * h'(x). Here, g(u) = u^(1/2) and h(x) = x + 1.

Differentiate g(u) = u^(1/2) to get g'(u) = (1/2)u^(-1/2). Differentiate h(x) = x + 1 to get h'(x) = 1. Therefore, f'(x) = (1/2)(x + 1)^(-1/2) * 1.

Evaluate the derivative at the given point x = 8 to find the slope of the tangent line. Substitute x = 8 into f'(x) to get f'(8) = (1/2)(8 + 1)^(-1/2).

Now, use the point-slope form of a line to find the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where m is the slope found in the previous step, and (x1, y1) is the given point (8, 3). Substitute these values into the equation to find the tangent line.

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Key Concepts

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

Derivative

The derivative of a function at a point provides the slope of the tangent line to the function's graph at that point. It is a fundamental concept in calculus that measures how a function changes as its input changes. For the function f(x) = √(x + 1), the derivative can be found using the power rule and chain rule.

Tangent Line

A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. It has the same slope as the curve at that point, which is given by the derivative. The equation of the tangent line can be found using the point-slope form of a line, y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point of tangency.

Point-Slope Form

The point-slope form of a linear equation is used to find the equation of a line when you know the slope and a point on the line. It is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a specific point on the line. This form is particularly useful for writing the equation of a tangent line once the slope is determined from the derivative.

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Slope-Intercept Form

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Textbook Question

Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = πh³. The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h.

Textbook Question

Tangent line to y = √x Does any tangent line to the curve y = √x cross the x-axis at x = −1? If so, find an equation for the line and the point of tangency. If not, why not?

Textbook Question

Differentiating Implicitly

Use implicit differentiation to find dy/dx in Exercises 1–14.

x³ + y³ = 18xy

Textbook Question

Differentiating Implicitly

Use implicit differentiation to find dy/dx in Exercises 1–14.

x²y + xy² = 6

Textbook Question

Derivatives

In Exercises 1–18, find dy/dx.

f(x) = x³ sin x cos x