/*! 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 32 Find \(y^{\prime \prime}\) $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(y^{\prime \prime}\) $$ y=\frac{3}{x^{4}}-\frac{1}{x} $$

Short Answer

Expert verified
The second derivative is \( y'' = 60x^{-6} - 2x^{-3} \).

Step by step solution

01

Recognize the Function Components

The given function is composed of several terms: \( y = \frac{3}{x^4} - \frac{1}{x} \). Express the function in terms of negative exponents for ease of differentiation: \( y = 3x^{-4} - x^{-1} \).
02

Differentiate to Find the First Derivative

Differentiate each term separately using the rule \(\frac{d}{dx}[x^{n}] = nx^{n-1}\). For the first term \(3x^{-4}\), the derivative is \(-12x^{-5}\). For the second term \(-x^{-1}\), the derivative is \(x^{-2}\). Thus, the first derivative, \(y'\), is \(-12x^{-5} + x^{-2}\).
03

Differentiate Again to Find the Second Derivative

Differentiate the first derivative \( y' = -12x^{-5} + x^{-2} \). For \(-12x^{-5}\), the derivative is \(60x^{-6}\). For \(x^{-2}\), the derivative is \(-2x^{-3}\). Therefore, the second derivative, \(y''\), is \(60x^{-6} - 2x^{-3}\).
04

Simplify the Second Derivative

Confirm that the expression for the second derivative is in its simplest form. Here, no further simplification is necessary, so the final expression for \(y''\) is \(60x^{-6} - 2x^{-3}\).

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Key Concepts

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

Differentiation Rules
In calculus, differentiation is all about measuring how a function changes as its input changes. There are some basic rules that make the process of finding the derivative easy, called differentiation rules. Here are a few key ones:
  • The Power Rule: This is a straightforward rule used to differentiate functions of the form \(x^n\). According to this rule, the derivative of \(x^n\) is \(nx^{n-1}\). This rule is especially useful when dealing with polynomial and power functions.
  • The Constant Rule: When differentiating a constant by itself, the derivative is always zero because a constant doesn’t change.
  • The Sum Rule: The derivative of a sum of functions is simply the sum of their derivatives. For example, if you have \(f(x) + g(x)\), the derivative will be \(f'(x) + g'(x)\).
By applying these rules, you can systematically find the derivative of complex functions by breaking them down into simpler parts.
Negative Exponents
Negative exponents can seem tricky, but they're really just a way to work with fractions in exponent format. A term with a negative exponent, like \(x^{-n}\), is the same as \(\frac{1}{x^n}\). This transformation is particularly helpful when differentiating, because it allows you to use the power rule easily.
This is exactly what we did with the original function \( y = \frac{3}{x^4} - \frac{1}{x} \). By rewriting it as \( y = 3x^{-4} - x^{-1} \), we converted fractions into expressions that could be directly differentiated using the power rule.
Working with negative exponents requires attentiveness, especially when simplifying expressions after differentiation. Always remember to apply the power rule carefully, which involves multiplying by the existing exponent and then subtracting one from it. This approach ensures consistency and correctness across your solutions.
Chain Rule
In differentiation, the chain rule is indispensable, especially when dealing with composite functions. The chain rule allows you to differentiate functions nested within each other, by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
Although the exercise you worked on didn't directly involve the chain rule, it's crucial when you face multi-layered functions, like \(f(g(x))\). Here's how it works in general terms:
  • If you have a composite function \(f(g(x))\), differentiate the outer function \(f\) with respect to \(g(x)\), and then multiply by the derivative of \(g(x)\), the inner function.
  • Symbolically, this looks like: \((f(g(x)))’ = f’(g(x)) \cdot g’(x)\).
The chain rule helps tackle complex differentiation tasks by breaking them down into smaller, manageable steps, making it a powerful tool in calculus.

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Most popular questions from this chapter

Then estimate the \(x\) -values at which tangent lines are horizontal. $$ f(x)=x^{4}-3 x^{2}+1 $$

Find the interval(s) for which \(f^{\prime}(x)\) is positive. $$ f(x)=x^{2}-4 x+1 $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow-\infty}\left(\frac{3 x^{3}+x+11}{6 x^{3}+x+2}\right) $$

The cost of sending a large envelope via U.S. first-class mail in 2014 was \(\$ 0.98\) for the first ounce and \(\$ 0.21\) for each additional ounce (or fraction thereof). (Source: www.usps.com.) If \(x\) represents the weight of a large envelope, in ounces, then \(p(x)\) is the cost of mailing it, where $$ \begin{array}{l} p(x)=\$ 0.98, \text { if } \quad 0 < x \leq 1, \\ p(x)=\$ 1.19, \text { if } \quad 1 < x \leq 2, \\ p(x)=\$ 1.40, \text { if } 2 < x \leq 3, \end{array} $$ and so on, up through 13 ounces. The graph of \(p\) is shown below. Using the graph of the postage function, find each of the following limits, if it exists. $$ \lim _{x \rightarrow 3.4} p(x) $$

Find the simplified difference quotient for each function listed. $$ f(x)=a x^{5}+b x^{4} $$

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