/*! 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 31 Find the first and second deriva... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the first and second derivatives of the functions in Exercises \(y=\frac{x^{3}+7}{x}\)

Short Answer

Expert verified
The first derivative is \( y' = 2x - 7x^{-2} \) and the second derivative is \( y'' = 2 + 14x^{-3} \).

Step by step solution

01

Rewrite the function

Start by rewriting the function to make it easier to differentiate. The function is given as \( y = \frac{x^3 + 7}{x} \). Separate the terms inside the fraction:\[y = \frac{x^3}{x} + \frac{7}{x} = x^2 + 7x^{-1}\]This will simplify our differentiation process.
02

Find the first derivative

Differentiate each term of the rewritten function with respect to \(x\). Using the power rule, \( \frac{d}{dx} (x^n) = nx^{n-1} \):1. The derivative of \( x^2 \) is \( 2x^{1} = 2x \).2. The derivative of \( 7x^{-1} \) is \( 7(-1)x^{-2} = -7x^{-2} \).Combine these results to find the first derivative:\[y' = 2x - 7x^{-2}\]
03

Find the second derivative

Differentiate the first derivative \( y' = 2x - 7x^{-2} \) once more to find the second derivative:1. The derivative of \( 2x \) is \( 2 \).2. The derivative of \( -7x^{-2} \) is \( -7(-2)x^{-3} = 14x^{-3} \).Combine these results to find the second derivative:\[y'' = 2 + 14x^{-3}\]

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Key Concepts

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

First Derivative
Calculus often involves finding derivatives, which tell us how a function changes at any given point. The first derivative of a function, often denoted as \( y' \) or \( \frac{dy}{dx} \), represents the slope of the tangent line to the function at a particular point. In simpler terms, it tells us the rate of change or the speed of something with respect to another variable.

For example, in the function \( y = \frac{x^3 + 7}{x} \), we started by rewriting the equation as \( y = x^2 + 7x^{-1} \) to make differentiation easier. By breaking it down into simpler terms, the power of each term becomes clear to apply the power rule.
  • The derivative of \( x^2 \) is \( 2x \), which indicates that for every unit increase in \( x \), \( y \) increases by \( 2x \).
  • For the term \( 7x^{-1} \), its derivative \( -7x^{-2} \) shows that as \( x \) increases, this part of the function decreases, affecting the overall rate of change.
Together, these derivatives combine to form the first derivative \( y' = 2x - 7x^{-2} \), showing us precisely how the function changes at any given value of \( x \). Understanding the first derivative is essential for analyzing functions, predicting trends, and conducting optimizations.
Second Derivative
Once we obtain the first derivative, we can explore the second derivative, which provides insight into the curvature or concavity of the original function. The second derivative, denoted as \( y'' \) or \( \frac{d^2y}{dx^2} \), describes how the rate of change itself changes. Think of it as the "acceleration" of the function.

From our exercise, the first derivative \( y' = 2x - 7x^{-2} \) was differentiated again to get the second derivative. Here's how:
  • The second derivative of \( 2x \) reduces to \( 2 \), indicating a constant slope.
  • For \( -7x^{-2} \), its derivative is \( 14x^{-3} \), implying that this term affects the change in curvature based on \( x \).
Putting them together, we determine that \( y'' = 2 + 14x^{-3} \). This provides an understanding of how the function behaves beyond the immediate slope, revealing its concave or convex nature. If \( y'' \) is positive at a point, \( y \) is concave up (think a "cup"), while a negative \( y'' \) implies a concave down form (shaped like an "umbrella"). This is pivotal in understanding points of inflection and behavior of the graph over time.
Power Rule
The power rule is a fundamental principle in calculus which simplifies the differentiation of polynomial functions. It states: if you have a term \( x^n \) in your function, its derivative is \( nx^{n-1} \). This rule makes differentiation straightforward for most standard forms.

In our exercise, where we had \( y = x^2 + 7x^{-1} \), both terms were perfect candidates for the power rule. Here's how it was applied:
  • For \( x^2 \), applying the power rule gives us: \( 2 \cdot x^{2-1} = 2x \).
  • For \( 7x^{-1} \), it's slightly tricky but still follows the rule: \( 7 \cdot (-1) \cdot x^{-1-1} = -7x^{-2} \).
The power rule streamlines the process of finding derivatives for polynomial terms by reducing repetitive calculations into a simple formula. By applying the power rule, you'll find derivatives with confidence and efficiency. Having this skill is crucial not only for calculus but for applying calculus concepts to physics, economics, and beyond.

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Most popular questions from this chapter

The derivative of \(\sin 2 x\) Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$ y=\frac{\sin 2(x+h)-\sin 2 x}{h} $$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h,\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

The linearization is the best linear approximation This is why we use the linearization.) 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 erroo \(E(x)=f(x)-g(x)\) 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 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) 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 \(x-a\)

Using the Chain Rule, show that the power rule \((d / d x) x^{n}=n x^{n-1}\) holds for the functions \(x^{n}\) in Exercises 109 and \(110 .\) $$ x^{3 / 4}=\sqrt{x \sqrt{x}} $$

In Exercises \(67-70\) , 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: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the lateral surface area \(S=2 \pi r h\) of a right circular cylinder when the height changes from \(h_{0}\) to \(h_{0}+d h\) and the radius does not change

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