Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 37b

Chapter 3, Problem 37b

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = 1/ √x; a= 1/4

Verified step by step guidance

Step 1: Find the derivative of the function f(x) = \(\frac{1}{\sqrt{x}\)}. Rewrite the function as f(x) = x^{-1/2} to make differentiation easier.

Step 2: Use the power rule for differentiation, which states that \(\frac{d}{dx}\)[x^n] = nx^{n-1}, to find f'(x). For f(x) = x^{-1/2}, the derivative f'(x) = -\(\frac{1}{2}\)x^{-3/2}.

Step 3: Evaluate the derivative at the given point a = \(\frac{1}{4}\). Substitute x = \(\frac{1}{4}\) into f'(x) to find the slope of the tangent line at this point.

Step 4: Calculate f(a) by substituting a = \(\frac{1}{4}\) into the original function f(x) = \(\frac{1}{\sqrt{x}\)}. This will give you the y-coordinate of the point of tangency.

Step 5: Use the point-slope form of a line, y - y_1 = m(x - x_1), where m is the slope found in Step 3 and (x_1, y_1) is the point (\(\frac{1}{4}\), f(\(\frac{1}{4}\))), to write the equation of the tangent line.

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 defined as the limit of the average rate of change of the function as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the graph of the function at that point.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. The equation of the tangent line can be expressed using the point-slope form, which incorporates the slope and the coordinates of the point of tangency.

Point-Slope Form

The point-slope form of a linear equation is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope (derivative) and the point of tangency are known. It allows for a straightforward way to express the line based on its slope and a specific point.

Recommended video:

Guided course

3:56

Slope-Intercept Form

Related Practice

Textbook Question

Calculate the derivative of the following functions.

y = sin (4x³ + 3x +1)

Textbook Question

Derivatives and tangent lines

a. For the following functions and values of a, find f′(a).

f(x) = 1/ x²; a= 1

Textbook Question

A feather dropped on the moon On the moon, a feather will fall to the ground at the same rate as a heavy stone. Suppose a feather is dropped from a height of 40 m above the surface of the moon. Its height (in meters) above the ground after t seconds is s = 40−0.8t². Determine the velocity and acceleration of the feather the moment it strikes the surface of the moon.

Textbook Question

Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a.

f(x) = √x+2; a=7

Textbook Question

Find the derivative function f' for the following functions f.

f(x) = √3x+1; a=8

views

Textbook Question

Calculate the derivative of the following functions.

y = csc (t² + t)

views