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Chapter 3: Problem 2

In Exercises 1鈥�12, find the first and second derivatives. $$ y=x^{2}+x+8 $$

Short Answer

Expert verified
First derivative: \( y' = 2x + 1 \). Second derivative: \( y'' = 2 \).

Step by step solution

01

Identify the Function

The function given is \( y = x^2 + x + 8 \). This is a polynomial function in terms of \( x \).
02

Differentiate to Find the First Derivative

To find the first derivative \( y' \), apply the power rule: \( \frac{d}{dx}[x^n]=nx^{n-1} \). For each term in the function:- For \( x^2 \), derivative is \( 2x \).- For \( x \), derivative is \( 1 \).- For constant \( 8 \), derivative is \( 0 \).Thus, the first derivative is \( y' = 2x + 1 \).
03

Differentiate Again to Find the Second Derivative

Differentiate the first derivative \( y' = 2x + 1 \) to get the second derivative \( y'' \):- The derivative of \( 2x \) is \( 2 \).- The derivative of the constant \( 1 \) is \( 0 \).Therefore, the second derivative is \( y'' = 2 \).

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Key Concepts

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

First Derivative
The first derivative of a function is crucial in understanding how the function behaves. It tells us how the function鈥檚 value changes as the variable changes, essentially giving us the slope or gradient of the function at any given point. This is helpful in determining where the function increases or decreases.
For the polynomial function \( y = x^2 + x + 8 \), we apply the **power rule**, which states \( \frac{d}{dx}[x^n]=nx^{n-1} \). Here's how it works for our function:
  • The derivative of \( x^2 \) is obtained by multiplying the exponent (2) by the coefficient (1), resulting in \( 2x^{2-1} = 2x \).
  • For the term \( x \), apply the power rule to get \( 1x^{1-1} = 1 \).
  • The constant 8 does not change as \( x \) changes, so its derivative is 0.
This leads to the first derivative \( y' = 2x + 1 \). Knowing this helps us analyze and predict changes in the graph of the function.
Second Derivative
The second derivative of a function gives insight into the curvature or concavity of the graph of the original function. It tells us how the rate of change itself is changing. In simpler terms, while the first derivative gives us the slope, the second derivative tells us how that slope changes.
To find the second derivative of the function \( y = x^2 + x + 8 \), we differentiate the first derivative \( y' = 2x + 1 \).
  • The derivative of \( 2x \) is 2. This is because the derivative of \( x \) with respect to \( x \) is \( 1 \), and calculated as \( 2 \times 1 = 2 \).
  • The constant 1's derivative is 0, as constants do not change.
Thus, the second derivative \( y'' = 2 \) indicates constant concavity. In this context, the positive constant indicates the function is concave up, meaning it curves upwards, and the slope continually increases.
Polynomial Functions
Polynomial functions are expressions composed of variables and coefficients that involve only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. A simple example is \( y = x^2 + x + 8 \), which is a quadratic polynomial.
  • Polynomial functions are key in calculus because their derivatives are very straightforward to calculate using the power rule.
  • They can model a wide variety of real-world situations where relationships are not perfectly linear but involve curvature.
  • Understanding the behavior of their graphs through derivatives is essential for predicting how changes in one variable affect another.
These functions are integral in both mathematical theory and practical applications such as physics, economics, and engineering, providing a functional way to represent and analyze curved relationships.

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Most popular questions from this chapter

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=2 \cos t, \quad y=2 \sin t, \quad t=\pi / 4 $$

Use a CAS to perform the following steps on the parametrized curves in Exercises \(113-116 .\) a. Plot the curve for the given interval of \(t\) values. b. Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(t_{0}\) . c. Find an equation for the tangent line to the curve at the point defined by the given value \(t_{0} .\) Plot the curve together with the tangent line on a single graph. $$ \begin{array}{l}{x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6} \\ {t_{0}=3 / 2}\end{array} $$

The diameter of a sphere is measured as \(100 \pm 1 \mathrm{cm}\) and the volume is calculated from this measurement. Estimate the percent- age error in the volume calculation.

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ x^{2}-\sqrt{3} x y+2 y^{2}=5, \quad(\sqrt{3}, 2) $$

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\cos t, \quad y=\sqrt{3} \cos t, \quad t=2 \pi / 3 $$
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