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In Exercises 1-12, find the first and second derivatives. \(y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{4}\)

Short Answer

Expert verified
First derivative: \(x^2 + x + \frac{1}{4}\); Second derivative: \(2x + 1\).

Step by step solution

01

Identify the Function Terms

We begin by identifying the components of the function.Given:\[ y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \]This function is composed of three polynomial terms: \(\frac{x^3}{3}\), \(\frac{x^2}{2}\), and \(\frac{x}{4}\).
02

Differentiate to Find the First Derivative

Using the power rule of differentiation, \(\frac{d}{dx} x^n = nx^{n-1}\), we find the first derivative of each term:- For \(\frac{x^3}{3}\): \[ \frac{d}{dx} \left(\frac{x^3}{3}\right) = \frac{3x^2}{3} = x^2 \]- For \(\frac{x^2}{2}\): \[ \frac{d}{dx} \left(\frac{x^2}{2}\right) = \frac{2x}{2} = x \]- For \(\frac{x}{4}\): \[ \frac{d}{dx} \left(\frac{x}{4}\right) = \frac{1}{4} \]Combine these results to find the first derivative:\[ y' = x^2 + x + \frac{1}{4} \]
03

Differentiate to Find the Second Derivative

Differentiate the first derivative to obtain the second derivative:- For \(x^2\): \[ \frac{d}{dx}(x^2) = 2x \]- For \(x\): \[ \frac{d}{dx}(x) = 1 \]- The constant \(\frac{1}{4}\) differentiates to 0.Combine these results to find the second derivative:\[ y'' = 2x + 1 \]

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Key Concepts

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

First Derivative
The first step in finding the first derivative is to understand the concept itself. A first derivative of a function gives us the rate at which the function is changing at any point. It is essentially the slope of the tangent line to the curve at any given point. For a given function \( y = f(x) \), the notation for the first derivative is usually \( y' \) or \( \frac{dy}{dx} \). The first derivative tells us whether the function is increasing or decreasing at each specific point.

In the given exercise, the function is \( y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \). Each term in the function needs to be differentiated independently using the power rule of differentiation, which simplifies the process.

Let's look closely at applying the power rule:
  • For \( \frac{x^3}{3} \): Differentiate to get \( x^2 \).
  • For \( \frac{x^2}{2} \): Differentiate to get \( x \).
  • For \( \frac{x}{4} \): Differentiate to yield a constant \( \frac{1}{4} \).
Adding these results provides the first derivative: \( y' = x^2 + x + \frac{1}{4} \). This derivative expression gives a simplified way to understand how the original function is changing at any given \( x \) value.
Second Derivative
The second derivative provides information about the curvature of the graph of the function, essentially the rate of change of the first derivative. It helps determine the concavity of a function: whether it is "concave up" (shaped like a cup \(\cup\)) or "concave down" (shaped like a cap \(\cap\)). The notation for the second derivative is \( y'' \) or \( \frac{d^2y}{dx^2} \).

In the example provided, the first derivative obtained is \( y' = x^2 + x + \frac{1}{4} \). This expression is then differentiated once more to find the second derivative. Using straightforward differentiation, we evaluate each term:
  • For \( x^2 \): Differentiate to get \( 2x \).
  • For \( x \): Differentiate to yield a constant \( 1 \).
  • The constant \( \frac{1}{4} \) differentiates to \( 0 \).
Combining these results gives the second derivative: \( y'' = 2x + 1 \). This function indicates the concavity of the original function; if \( y'' > 0 \), the graph is concave up. If \( y'' < 0 \), the graph is concave down at that point.
Power Rule of Differentiation
The power rule of differentiation is one of the simplest and most commonly used rules in calculus, allowing for the quick differentiation of polynomial functions. The rule states that for any power function of the form \( x^n \), the derivative with respect to \( x \) is \( nx^{n-1} \). This means you multiply by the power, and then reduce the power by one.

The power rule is extremely useful when dealing with polynomial terms:
  • When you have \( x^3 \), the derivative using the power rule is \( 3x^2 \).
  • For \( x^2 \), the derivative becomes \( 2x \).
  • Even a simple \( x \) is differentiated to \( 1 \), as \( x^{1-1} = x^0 = 1 \).
It’s important to understand that this rule applies to each term individually, making it straightforward to find derivatives for sums and differences of polynomial expressions. In our example, this rule enabled quick and accurate differentiation of each component of the function, leading to the first and second derivatives we discussed.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(53-60 .\) \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x^{3}-x y+y^{3}=7, \quad P(2,1) \end{equation}

If \(L=\sqrt{x^{2}+y^{2}}, d x / d t=-1,\) and \(d y / d t=3,\) find \(d L / d t\) when \(x=5\) and \(y=12 .\)

A draining conical reservoir Water is flowing at the rate of 50 \(\mathrm{m}^{3} / \mathrm{min}\) from a shallow concrete conical reservoir (vertex down) of base radius 45 \(\mathrm{m}\) and height 6 \(\mathrm{m} .\) a. How fast (centimeters per minute) is the water level falling when the water is 5 \(\mathrm{m}\) deep? b. How fast is the radius of the water's surface changing then? Answer in centimeters per minute.

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1 $$

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

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