Problem 47
Question
Find \(y^{\prime}\) $$ y=\frac{x}{7}+\frac{7}{x} $$
Step-by-Step Solution
Verified Answer
\( y' = \frac{1}{7} - \frac{7}{x^2} \)
1Step 1: Differentiate the first term
The expression given is \( y = \frac{x}{7} + \frac{7}{x} \). To find \( y' \), we differentiate each term individually. Begin with the term \( \frac{x}{7} \). Notice this is equivalent to \( \frac{1}{7}x \). Therefore, the derivative is \( \frac{1}{7} \cdot 1 = \frac{1}{7} \).
2Step 2: Differentiate the second term
Now, differentiate the second term, \( \frac{7}{x} \). This can be rewritten using a negative exponent as \( 7x^{-1} \). The derivative of \( 7x^{-1} \) is \( 7 \cdot (-1)x^{-2} = -\frac{7}{x^2} \).
3Step 3: Combine the derivatives
Combine the results from Step 1 and Step 2 to find the derivative of the entire expression: \( y' = \frac{1}{7} - \frac{7}{x^2} \).
Key Concepts
Derivative of a functionPower ruleNegative exponent
Derivative of a function
In calculus, the derivative of a function helps us understand how the function behaves as its input changes. It provides the rate at which the function value changes relative to changes in its input.
To determine the derivative, we use differentiation techniques. Differentiating allows us to find the slope of the tangent line at any point on a curve described by a function. This is essential in fields like physics and engineering, where understanding change is crucial.
In simple terms, if you have a function represented as \( y = f(x) \), its derivative, often written as \( y' \) or \( f'(x) \), measures how \( y \) changes with respect to \( x \). In our exercise, we differentiate each term of \( y = \frac{x}{7} + \frac{7}{x} \) separately to find \( y' \). This process involves applying rules, such as the power rule, to find the correct rate of change for each part of the function.
To determine the derivative, we use differentiation techniques. Differentiating allows us to find the slope of the tangent line at any point on a curve described by a function. This is essential in fields like physics and engineering, where understanding change is crucial.
In simple terms, if you have a function represented as \( y = f(x) \), its derivative, often written as \( y' \) or \( f'(x) \), measures how \( y \) changes with respect to \( x \). In our exercise, we differentiate each term of \( y = \frac{x}{7} + \frac{7}{x} \) separately to find \( y' \). This process involves applying rules, such as the power rule, to find the correct rate of change for each part of the function.
Power rule
The power rule is a fundamental tool in calculus for finding derivatives. It states that if you have a term of the form \( x^n \), its derivative is \( nx^{n-1} \).
This simple yet powerful rule makes it easy to differentiate polynomial expressions, where each term can be treated individually. For example, if you have \( x^3 \), applying the power rule gives \( 3x^2 \), since the exponent 3 moves in front of \( x \) and is reduced by one.
In our specific exercise, when differentiating \( \frac{x}{7} \), which is effectively \( \frac{1}{7}x \), we treat \( x \) as \( x^1 \). By applying the power rule, the derivative of \( \frac{1}{7}x^1 \) becomes \( \frac{1}{7} \cdot 1 \cdot x^{1-1} = \frac{1}{7} \). This illustrates how the power rule simplifies the process of differentiation.
This simple yet powerful rule makes it easy to differentiate polynomial expressions, where each term can be treated individually. For example, if you have \( x^3 \), applying the power rule gives \( 3x^2 \), since the exponent 3 moves in front of \( x \) and is reduced by one.
In our specific exercise, when differentiating \( \frac{x}{7} \), which is effectively \( \frac{1}{7}x \), we treat \( x \) as \( x^1 \). By applying the power rule, the derivative of \( \frac{1}{7}x^1 \) becomes \( \frac{1}{7} \cdot 1 \cdot x^{1-1} = \frac{1}{7} \). This illustrates how the power rule simplifies the process of differentiation.
Negative exponent
Working with negative exponents is a helpful technique when differentiating functions that involve terms in a fraction form. A negative exponent indicates that the base is on the other side of a fraction.
For example, any term \( x^{-n} \) is equivalent to \( \frac{1}{x^n} \). In differentiation, this approach simplifies the application of the power rule to expressions that are initially fractions.
In the original exercise, the term \( \frac{7}{x} \) is rewritten as \( 7x^{-1} \). This allows us to use the power rule effectively. Differentiating \( 7x^{-1} \) with respect to \( x \) by applying the power rule gives us \( 7 \cdot (-1)x^{-1-1} = -7x^{-2} \), which is \( -\frac{7}{x^2} \). Understanding how to work with negative exponents makes it easier to handle a wide range of differentiation challenges.
For example, any term \( x^{-n} \) is equivalent to \( \frac{1}{x^n} \). In differentiation, this approach simplifies the application of the power rule to expressions that are initially fractions.
In the original exercise, the term \( \frac{7}{x} \) is rewritten as \( 7x^{-1} \). This allows us to use the power rule effectively. Differentiating \( 7x^{-1} \) with respect to \( x \) by applying the power rule gives us \( 7 \cdot (-1)x^{-1-1} = -7x^{-2} \), which is \( -\frac{7}{x^2} \). Understanding how to work with negative exponents makes it easier to handle a wide range of differentiation challenges.
Other exercises in this chapter
Problem 46
Differentiate each function. \(f(x)=\frac{3 x^{2}+2 x}{x^{2}+1}\)
View solution Problem 47
Find the simplified difference quotient for each function listed. $$ f(x)=a x^{2}+b x+c $$
View solution Problem 47
Find \(f^{\prime}(x)\) for the given function. \(f(x)=\frac{1}{\sqrt{x}}(\) See Exercise 56 in Section \(1.3 .)\)
View solution Problem 47
Given $$ s(t)=3 t+10 $$ where \(s(t)\) is in miles and \(t\) is in hours, find each of the following. a) \(v(t)\) b) \(a(t)\) c) The velocity and acceleration w
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