Problem 47
Question
Consider the following functions and express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(d y=f^{\prime}(x) d x\). \(f(x)=3 x^{3}-4 x\)
Step-by-Step Solution
Verified Answer
Question: Express the relationship between a small change in x and the corresponding change in y for the function \(f(x) = 3x^3 - 4x\).
Answer: For the function \(f(x) = 3x^3 - 4x\), the relationship between a small change in x (represented by \(dx\)) and the corresponding change in y (represented by \(dy\)) is given by the equation \(dy = (9x^2 - 4)dx\).
1Step 1: Differentiate the given function
To find the rate of change of y with respect to x, we need to differentiate the given function \(f(x) = 3x^3 - 4x\) with respect to x. By applying the power rule and sum/difference rule of differentiation, we get:
\(f'(x) = \frac{d}{dx}(3x^3 - 4x) = 3\frac{d}{dx}(x^3) - 4\frac{d}{dx}(x)\)
\(f'(x) = 3(3x^2) - 4(1)\)
\(f'(x) = 9x^2 - 4\)
Now that we have the derivative of the function \(f'(x)\), we can move to the next step.
2Step 2: Express the relationship between small changes in x and y
We can use the differential form \(dy = f'(x)dx\) to express the relationship between the small change in x (represented by \(dx\)) and the corresponding change in y (represented by \(dy\)). With the derivative we found in the previous step, we have:
\(dy = (9x^2 - 4)dx\)
This equation represents the relationship between a small change in x and the corresponding change in y for the function \(f(x) = 3x^3 - 4x\).
Key Concepts
Power RuleSum/Difference RuleRate of Change
Power Rule
The power rule is a fundamental tool in calculus used for differentiating functions of the form \(x^n\), where \(n\) is any real number. It provides a straightforward way to find the derivative of polynomial expressions. To apply the power rule, you first multiply the term by its exponent and then subtract one from the exponent.
For example, if you have a term \(x^n\), its derivative using the power rule is \(nx^{n-1}\). In the given exercise, the function \(f(x) = 3x^3 - 4x\) applies the power rule as follows:
Remember, when using the power rule, each term in the function is treated independently, and you simply apply the same procedure to each term.
For example, if you have a term \(x^n\), its derivative using the power rule is \(nx^{n-1}\). In the given exercise, the function \(f(x) = 3x^3 - 4x\) applies the power rule as follows:
- The term \(3x^3\) becomes \(3 \times 3x^{2} = 9x^2\).
- The term \(-4x\) becomes \(-4 \times 1x^{0} = -4\), since \(x\) is considered as \(x^1\).
Remember, when using the power rule, each term in the function is treated independently, and you simply apply the same procedure to each term.
Sum/Difference Rule
The sum/difference rule in differentiation allows you to differentiate each term of a sum or difference in a function separately. This rule is particularly useful in handling polynomial functions where terms are added or subtracted. According to the sum/difference rule, you apply differentiation to each individual term separately and then combine the results.
In our example, the function is \(f(x) = 3x^3 - 4x\). This expression can be seen as the sum of two terms: \(3x^3\) and \(-4x\). By using the sum/difference rule, you differentiate as follows:
In our example, the function is \(f(x) = 3x^3 - 4x\). This expression can be seen as the sum of two terms: \(3x^3\) and \(-4x\). By using the sum/difference rule, you differentiate as follows:
- Differentiate \(3x^3\) using the power rule: result is \(9x^2\).
- Differentiate \(-4x\) using the power rule: result is \(-4\).
Rate of Change
In calculus, the concept of "rate of change" describes how a quantity changes in relation to another quantity. It's essentially what we measure when we differentiate a function. The derivative \(f'(x)\) paints a clear picture of how \(f(x)\), changes as \(x\) changes.
With our function \(f(x) = 3x^3 - 4x\), once differentiated, the rate of change is expressed as \(f'(x) = 9x^2 - 4\). This derivative tells us the rate at which \(y\) (the outcome of the function) changes as \(x\) changes.
This concept is applied using the differential form \(dy = f'(x)dx\), meaning any small change in \(x\) will correspond to a change in \(y\). For the given function, this relationship is articulated by \(dy = (9x^2 - 4)dx\), showing the instantaneous rate of change at any point \(x\).
With our function \(f(x) = 3x^3 - 4x\), once differentiated, the rate of change is expressed as \(f'(x) = 9x^2 - 4\). This derivative tells us the rate at which \(y\) (the outcome of the function) changes as \(x\) changes.
This concept is applied using the differential form \(dy = f'(x)dx\), meaning any small change in \(x\) will correspond to a change in \(y\). For the given function, this relationship is articulated by \(dy = (9x^2 - 4)dx\), showing the instantaneous rate of change at any point \(x\).
- "Instantaneous" means no matter how small \(dx\) is, this rate remains consistent for that small interval.
- It's valuable in scenarios where you need to understand dynamics, like physics or economics.
Other exercises in this chapter
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