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Find the first and second derivatives of the functions in Exercises 33-40. \(y=\frac{x^{3}+7}{x}\)

Short Answer

Expert verified
First derivative: \( y' = 2x - \frac{7}{x^2} \); Second derivative: \( y'' = 2 + \frac{14}{x^3} \).

Step by step solution

01

Rewrite the Function

We start with the function \( y = \frac{x^3 + 7}{x} \). To simplify differentiation, we'll rewrite the function by dividing each term in the numerator by the denominator. Thus, we have:\[ y = \frac{x^3}{x} + \frac{7}{x} = x^2 + 7x^{-1} \]
02

Find the First Derivative

Using the rewritten function \( y = x^2 + 7x^{-1} \), we'll differentiate term by term.For \( x^2 \): Apply the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), which gives us \( 2x^{2-1} = 2x \).For \( 7x^{-1} \): Apply the power rule again, resulting in \( 7(-1)x^{-2} = -7x^{-2} \).Therefore, the first derivative is:\[ y' = 2x - \frac{7}{x^2} \]
03

Find the Second Derivative

Now, we need to find the derivative of the first derivative \( y' = 2x - \frac{7}{x^2} \).For \( 2x \): Differentiate using the power rule to get \( 2 \).For \( -\frac{7}{x^2} \): Rewrite it as \( -7x^{-2} \) and differentiate to get \(-7(-2)x^{-3} = 14x^{-3} = \frac{14}{x^3} \).Thus, the second derivative is:\[ y'' = 2 + \frac{14}{x^3} \]

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Key Concepts

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

Understanding the First Derivative
The process of finding the first derivative is all about acquiring the rate of change or the slope of a function at any given point. When dealing with functions, the first derivative provides insight into whether the function is increasing or decreasing.
In our example, we start with the function \( y = \frac{x^3 + 7}{x} \). To simplify the differentiation process, we rewrite it as \( y = x^2 + 7x^{-1} \).

The power rule is the key technique used here, which states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule allows us to easily differentiate each term separately in the function.
  • For \( x^2 \): Applying the power rule, we find the derivative to be \( 2x^{2-1} = 2x \).
  • For \( 7x^{-1} \): Using the power rule again, we derive \( 7(-1)x^{-2} = -7x^{-2} \).
Bringing these parts together, the first derivative thus is:
\[ y' = 2x - \frac{7}{x^2} \] This expression gives us the slope of the tangent line to the graph of the function at any specific value of \( x \).
Exploring the Second Derivative
The second derivative offers us further insight into the curvature or concavity of a function's graph. Simply put, it tells us how the rate of change is changing. If positive, the function is concave up; if negative, concave down.
To find the second derivative, we differentiate the first derivative that we previously found.
Starting with the first derivative:
\[ y' = 2x - \frac{7}{x^2} \]
  • For \( 2x \): The derivative using the power rule simply gives us a constant \( 2 \).
  • For \( -\frac{7}{x^2} \): We rewrite it as \( -7x^{-2} \) and apply the power rule to get \(-7(-2)x^{-3} = 14x^{-3} \), which simplifies to \( \frac{14}{x^3} \).
Putting all components together, the second derivative is:
\[ y'' = 2 + \frac{14}{x^3} \]
This derivative helps in understanding the changes in the slope of the function. If \( y'' > 0 \), it indicates that the function is concave up, reflecting a minimum point. Conversely, if \( y'' < 0 \), it indicates a concave down situation, reflecting a maximum point.
Mastering the Power Rule
The power rule is one of the most fundamental and frequently used rules in calculus differentiation. It simplifies finding derivatives for functions in the form of \( x^n \), where \( n \) is any real number.
According to the rule, \( \frac{d}{dx} x^n = nx^{n-1} \). This simple formula allows us to determine the derivative of power functions quickly and easily.
  • It is applicable to positive, negative, or even fractional exponents, which makes it incredibly versatile.
  • For a linear term like \( x^2 \), the power rule gives a straightforward derivative of \( 2x \).
  • When dealing with terms like \( 7x^{-1} \), the rule readily handles the negative exponent, simplifying differentiation processes to \( -7x^{-2} \).
The power rule thus forms the backbone of many differentiation problems and is essential for mastering calculus differentiation techniques. Understanding and applying this rule effectively enables students to tackle a wide range of problems with ease and confidence.

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