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Chapter 1: Problem 11

Find the first and second derivatives. \(y=x+1\)

Short Answer

Expert verified
The first derivative is 1 and the second derivative is 0.

Step by step solution

01

- Find the First Derivative

Differentiate the given function with respect to x. The given function is y = x + 1.
02

- Apply the Power Rule

Using the power rule for each term separately. For the term x, the derivative is 1 because the power rule states that y' = nx^{n-1}. Here n=1, so it becomes 1. The derivative of a constant term is 0. Therefore, y' = 1 + 0 = 1.
03

- Find the Second Derivative

Differentiate the first derivative with respect to x. The first derivative y' = 1. Applying the power rule again, y'' = 0 because the derivative of a constant is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
In calculus, finding the 'first derivative' of a function is extremely important. The first derivative, often represented as y' or \(\frac{dy}{dx}\), tells us the rate at which the function is changing at any point. It's like finding the slope of the function at any given x-value.

For the given exercise, we start with the function y = x + 1. To find the first derivative, we differentiate this function with respect to x. Here鈥檚 how:

  • Differentiating each term separately: Since the function is composed of two terms, x and 1, we apply the derivative rule to each:

The derivative of x is 1, as per the power rule, where \(nx^{n-1}\) becomes 1 when n is 1.

The derivative of the constant term 1 is 0.

Putting it all together, the first derivative y' is 1.
Second Derivative
The 'second derivative' provides insights into the curvature and concavity of the function. Represented by y'' or \(\frac{d^2y}{dx^2}\), it helps us understand whether the function is concave up or down at a particular point.

For our function, where we already found the first derivative y' = 1, let鈥檚 now find the second derivative:

We differentiate the first derivative with respect to x.

Since y' = 1, and 1 is a constant, the derivative of a constant is always 0.

Thus, the second derivative y'' = 0. This indicates that the rate of change of the slope is zero, meaning a straight line with no curvature.
Power Rule
The 'power rule' is one of the most commonly used rules in differentiation. It allows us to easily find the derivative of functions of the form \(x^n\). Here鈥檚 a breakdown of how it works:

The power rule states: if y = \(x^n\), then the derivative, y' = \(nx^{n-1}\).

In simpler terms, you multiply the coefficient by the exponent (n), and then subtract 1 from the exponent.

For our exercise, let鈥檚 see how the power rule was applied:
  • For the term x in the function y = x + 1, which is the same as \(x^1\), n is 1.
  • Using the rule, the derivative of x is 1*\

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Most popular questions from this chapter

Let \(R(x)\) denote the revenue (in thousands of dollars) generated from the production of \(x\) units of computer chips per day, where each unit consists of 100 chips. (a) Represent the following statement by equations involving \(R\) or \(R^{\prime}\) : When 1200 chips are produced per day, the revenue is \(\$ 22,000\) and the marginal revenue is \(\$ .75\) per chip. (b) If the marginal cost of producing 1200 chips is \(\$ 1.5\) per chip, what is the marginal profit at this production level?

Find the first and second derivatives. \(y=100\)

The functions in Exercises 21-26 are defined for all \(x\) except for one value of \(x\). If possible, define \(f(x)\) at the exceptional point in a way that makes \(f(x)\) continuous for all \(x\). \(f(x)=\frac{\sqrt{9+x}-\sqrt{9}}{x}, x \neq 0\)

Sale of T-Shirts A T-shirt shop owner has a fixed cost of \(\$ 230\) and a marginal cost of \(\$ 7\) per T-shirt to manufacture \(x\) T-shirts per day. Let \(C(x)\) denote the cost to manufacture \(x\) T-shirts per day. (a) Find \(C(x)\). (b) If the shop owner decides to sell the T-shirts at \(\$ 12\) each, find \(R(x)\), the total revenue from selling \(x\) T-shirts per day.

A helicopter is rising straight up in the air. Its distance from the ground \(t\) seconds after takeoff is \(s(t)\) feet, where \(s(t)=t^{2}+t\). (a) How long will it take for the helicopter to rise 20 feet? (b) Find the velocity and the acceleration of the helicopter when it is 20 feet above the ground.
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