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Chapter 3: Problem 29

Find the derivatives of all orders of the functions in Exercises 29-32. \(y=\frac{x^{4}}{2}-\frac{3}{2} x^{2}-x\)

Short Answer

Expert verified
First derivative: \(2x^3 - 3x - 1\); second: \(6x^2 - 3\); third: \(12x\); fourth: \(12\); fifth and higher: \(0\).

Step by step solution

01

Find the First Derivative

To find the first derivative of the function \(y = \frac{x^4}{2} - \frac{3}{2}x^2 - x\), apply the power rule which states that the derivative of \(x^n\) is \(nx^{n-1}\). Thus: \[y' = \frac{d}{dx}\left(\frac{x^4}{2}\right) - \frac{d}{dx}\left(\frac{3}{2}x^2\right) - \frac{d}{dx}(x)\]Calculate each term separately: - \(\frac{d}{dx}\left(\frac{x^4}{2}\right) = \frac{4}{2}x^{3} = 2x^{3}\)- \(\frac{d}{dx}\left(\frac{3}{2}x^{2}\right) = \frac{3 \cdot 2}{2}x^{1} = 3x\)- \(\frac{d}{dx}(x) = 1\)Combine these: \[y' = 2x^{3} - 3x - 1\]
02

Find the Second Derivative

To find the second derivative, take the derivative of the first derivative \(y' = 2x^{3} - 3x - 1\). \[y'' = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x) - \frac{d}{dx}(1)\]Calculate each term: - \(\frac{d}{dx}(2x^3) = 6x^{2}\)- \(\frac{d}{dx}(3x) = 3\)- \(\frac{d}{dx}(1) = 0\)Therefore, combine them to find \[y'' = 6x^{2} - 3\]
03

Find the Third Derivative

To find the third derivative, differentiate the second derivative \(y'' = 6x^{2} - 3\).\[y''' = \frac{d}{dx}(6x^2) - \frac{d}{dx}(3)\]Calculate each term: - \(\frac{d}{dx}(6x^2) = 12x\)Thus, \[y''' = 12x\]
04

Find the Fourth Derivative

To find the fourth derivative, differentiate the third derivative \(y''' = 12x\).\[y^{(4)} = \frac{d}{dx}(12x)\]Calculate the derivative: - \(\frac{d}{dx}(12x) = 12\)So, the fourth derivative is \[y^{(4)} = 12\]
05

Find Higher Order Derivatives

Observe that the derivative of a constant (which is the fourth derivative we calculated, 12) is zero. Thus, the fifth and higher-order derivatives will be zero.Therefore, \[y^{(5)} = y^{(6)} = \ldots = 0\]

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Key Concepts

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

Power Rule
The power rule is one of the most fundamental rules in calculus, used for differentiating functions of the form \( x^n \). The rule simply states that the derivative of \( x^n \) is \( nx^{n-1} \). This means you bring the exponent down in front of the \( x \) and then reduce the exponent by one. For instance:
  • If you have \( x^4 \), applying the power rule gives \( 4x^3 \).
  • For \( \frac{x^4}{2} \), it becomes \( \frac{4}{2}x^3 = 2x^3 \).
This straightforward mechanism allows you to differentiate polynomial terms quickly. It is a core tool in solving differentials, frequently utilized in calculus to simplify expressions and analyze functions.
Higher Order Derivatives
Once you find the first derivative, you can keep differentiating the function to find higher order derivatives. These derivatives are used to understand the behavior of the function at a deeper level. Each successive derivative tells you more about the function's change:
  • The first derivative indicates the slope or the rate of change of the function.
  • Higher order derivatives, like the second, third, or fourth, can show concavity and the rate at which the slope itself is changing.
In our exercise, the fourth derivative turned out to be a constant (12), with all higher derivatives becoming zero. This illustrates that a polynomial eventually leads to constant or zero derivatives.
First Derivative
The first derivative of a function describes its rate of change or the slope of the tangent line at any point. For a function \( y = \frac{x^4}{2} - \frac{3}{2}x^2 - x \), finding the first derivative involves applying the power rule to each term:- The term \( \frac{x^4}{2} \) becomes \( 2x^3 \), since \( \frac{4}{2} \times x^{3} = 2x^3 \).- For \( -\frac{3}{2}x^2 \), it converts to \( -3x \) because \( \frac{3 \cdot 2}{2} \times x^{1} = 3x \).- The derivative of \( -x \) is simply \( -1 \).Thus, the first derivative is \( y' = 2x^3 - 3x - 1 \). This equation helps you determine how the function behaves at any given value of \( x \).
Second Derivative
The second derivative provides insight into the concavity of the function. It tells you whether the slope of the first derivative is increasing or decreasing. In the given exercise, we take the first derivative equation \( y' = 2x^3 - 3x - 1 \) and differentiate it again:- \( \, \frac{d}{dx}(2x^3) = 6x^2 \, \), implying the slope of this term is changing at twice the original rate.- \( \, \frac{d}{dx}(-3x) = -3 \, \) means the slope is constant and negative.- The constant \( \, -1 \, \) has a derivative of zero.Therefore, the second derivative is \( y'' = 6x^2 - 3 \). Analyzing this, if \( y'' > 0 \), the function is concave up at that interval, and if \( y'' < 0 \), it is concave down. This is crucial for understanding the nature and extremes of the curve.

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Most popular questions from this chapter

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the surface area } S=6 x^{2} \text { of a cube when the edge }} \\ {\text { lengths change from } x_{0} \text { to } x_{0}+d x}\end{array} $$

Estimating height of a building A surveyor, standing 30 \(\mathrm{ft}\) from the base of a building, measures the angle of elevation to the top of the building to be \(75^{\circ} .\) How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4\(\% ?\)

Use a CAS to perform the following steps. \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+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) \end{equation}

The linearization is the best linear approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions $$ \begin{array}{l}{\text { 1. } E(a)=0} \\ {\text { 2. } \lim _{x \rightarrow a} \frac{E(x)}{x-a}=0}\end{array} $$ then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\) .

Tolerance The height and radius of a right circular cylinder are equal, so the cylinder's volume is \(V=\pi h^{3} .\) The volume is to be calculated with an error of no more than 1\(\%\) of the true Find approximately the greatest error that can be tolerated in the measurement of \(h\) , expressed as a percentage of \(h .\)
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