/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 Find the first and the second de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the first and the second derivatives of each function. \(f(x)=\frac{1}{x^{2}}+x-x^{3}\)

Short Answer

Expert verified
The first derivative is \( f'(x) = -2x^{-3} + 1 - 3x^2 \) and the second derivative is \( f''(x) = 6x^{-4} - 6x \).

Step by step solution

01

Rewrite the Function

Before differentiating, rewrite the function in a more convenient form: \[ f(x) = x^{-2} + x - x^3 \]. This allows us to apply power rule directly for differentiation.
02

Differentiate to Find the First Derivative

Apply the power rule \( \frac{d}{dx} x^n = n x^{n-1} \) to each term in \( f(x) = x^{-2} + x - x^3 \):- The derivative of \( x^{-2} \) is \( -2x^{-3} \).- The derivative of \( x \) is \( 1 \).- The derivative of \( -x^3 \) is \( -3x^2 \).Thus, the first derivative is:\[ f'(x) = -2x^{-3} + 1 - 3x^2 \].
03

Differentiate Again to Find the Second Derivative

Differentiate \( f'(x) = -2x^{-3} + 1 - 3x^2 \) using the power rule again:- The derivative of \( -2x^{-3} \) is \( 6x^{-4} \).- The derivative of \( 1 \) is \( 0 \).- The derivative of \( -3x^2 \) is \( -6x \).Therefore, the second derivative is:\[ f''(x) = 6x^{-4} - 6x \].

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Key Concepts

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

Understanding the Power Rule
The power rule is a fundamental tool for calculating derivatives in calculus. It is particularly useful for functions expressed as powers of a variable. The rule itself states that if you have a function of the form \( f(x) = x^n \), then its derivative can be found using the formula \( \frac{d}{dx} x^n = nx^{n-1} \). This concept makes it incredibly easy to differentiate terms that involve powers of \( x \).

In our exercise, we first rewrote the function \( f(x) = \frac{1}{x^2} + x - x^3 \) in terms of negative and positive powers: \( f(x) = x^{-2} + x - x^3 \). This rewriting helps utilize the power rule effectively for each term:
  • For \( x^{-2} \), applying the power rule gives the derivative \( -2x^{-3} \).
  • For \( x \), which is really \( x^1 \), the derivative is simply \( 1 \times x^{0} = 1 \).
  • For \( -x^3 \), we obtain \( -3x^2 \) as the derivative.
This simplification makes taking derivatives straightforward and reduces the risk of error.
Finding the First Derivative
The first derivative of a function measures how the function's output changes as the input changes. That is, it tells us the rate of change or the slope of the function at any given point. To find the first derivative, we apply the power rule to each term of our rewritten function:

- \( x^{-2} \) becomes \( -2x^{-3} \)
- \( x \) becomes \( 1 \)
- \( -x^3 \) becomes \( -3x^2 \)

Adding these, the first derivative of our function \( f(x) = x^{-2} + x - x^3 \) is:
\[ f'(x) = -2x^{-3} + 1 - 3x^2 \]

This result provides a new function that describes the slope of the original function at every point along its graph. Understanding the first derivative is crucial as it helps to reveal where the function is increasing, decreasing, or has any constant segments.
Exploring the Second Derivative
The second derivative gives us a deeper understanding of the function's behavior by showing us the rate at which the slope is changing. In practical terms, it's the derivative of the derivative. It can reveal whether a function is concave or convex, or identify points of inflection.

To find the second derivative, we differentiate the first derivative \( f'(x) = -2x^{-3} + 1 - 3x^2 \) again. Using the power rule once more for each term results in:
  • The derivative of \( -2x^{-3} \) is \( 6x^{-4} \).
  • The derivative of the constant \( 1 \) is \( 0 \), because constants do not change as \( x \) changes.
  • The derivative of \( -3x^2 \) is \( -6x \).
Thus, the second derivative is:
\[ f''(x) = 6x^{-4} - 6x \]
This second derivative informs us about the concavity of the original function \( f(x) \). If the second derivative is positive, the function is concave up (like a cup); if negative, concave down (like a frown). It helps in sketching curves and analyzing how a function behaves beyond just slope changes.

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Most popular questions from this chapter

Suppose that you are studying reproduction in moss. The scaling relation $$N \propto L^{2.11}$$ has been found (Niklas, 1994 ) between the number of moss spores \((N)\) and the capsule length \((L)\). This relation is not very accurate, but it turns out that it suffices for your purpose. To estimate the number of moss spores, you measure the capsule length. If you wish to estimate the number of moss spores within an error of \(5 \%\), how accurately must you measure the capsule length?

Assume that the measurement of \(x\) is accurate within \(2 \% .\) In each case, determine the error \(\Delta f\) in the calculation of \(f\) and find the percentage error \(100 \frac{\Delta f}{f} .\) The quantities \(f(x)\) and the true value of \(x\) are given. $$ f(x)=\frac{1}{1+x}, x=4 $$

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. $$ \sin \left(\frac{\pi}{2}+0.02\right) $$

Suppose the function \(f(x)\) is piecewise defined; that is, \(f(x)=f_{1}(x)\) for \(x \leq a\) and \(f(x)=f_{2}(x)\) for \(x>a\). Assume that \(f_{1}(x)\) is continuous and differentiable for \(xa\). Sketch graphs of \(f(x)\) for the following three cases: (a) \(f(x)\) is continuous and differentiable at \(x=a\). (b) \(f(x)\) is continuous, but not differentiable, at \(x=a\). (c) \(f(x)\) is neither continuous nor differentiable at \(x=a\).

Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.

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