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Suppose \(u\) and \(v\) are functions with a common domain and \(P=u-v\). Write \(P(t)=u(t)+(-1) v(t)\) and use the Sum and Constant Factor rules to show that \(P^{\prime}(t)=u^{\prime}(t)-v^{\prime}(t)\)

Short Answer

Expert verified
\(P'(t) = u'(t) - v'(t)\)

Step by step solution

01

Rewrite the Function P

We start with the function \(P(t) = u(t) - v(t)\). This can be rewritten as \(P(t) = u(t) + (-1) \cdot v(t)\). This step sets up the functions to use the Sum and Constant Factor rules.
02

Apply the Sum Rule

According to the Sum Rule, if you have a function that is the sum of two other functions \(f(t) = g(t) + h(t)\), then the derivative \(f'(t) = g'(t) + h'(t)\). Applying this to \(P(t) = u(t) + (-1) \cdot v(t)\), we have \(P'(t) = u'(t) + (d/dt)((-1) \cdot v(t))\).
03

Apply the Constant Factor Rule

The Constant Factor Rule states that the derivative of a constant multiplied by a function \(k \cdot f(t)\) is \(k \cdot f'(t)\). Applying this to \((-1) \cdot v(t)\) yields \((d/dt)((-1) \cdot v(t)) = -1 \cdot v'(t)\).
04

Combine Results

Substitute the results from the Constant Factor Rule into the result of the Sum Rule: \(P'(t) = u'(t) + (-1) \cdot v'(t)\). This simplifies to \(P'(t) = u'(t) - v'(t)\). Therefore, we have shown that \(P'(t) = u'(t) - v'(t)\).

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Key Concepts

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

Sum Rule
The Sum Rule is a fundamental concept in calculus, particularly when dealing with derivatives of functions. The idea is straightforward: If you have two functions, say \(g(t)\) and \(h(t)\), that are added together to form a new function, \(f(t) = g(t) + h(t)\), then the derivative of this new function is simply the sum of the derivatives of the original functions. In mathematical terms, this is expressed as:\[f'(t) = g'(t) + h'(t) \]This rule makes it easy to handle derivatives of more complex expressions without having to break them down into smaller parts repeatedly. It's an example of how calculus simplifies the process of finding rates of change, by allowing you to apply properties to combinations of functions.In the exercise, the function \(P(t) = u(t) - v(t)\) was restated as \(P(t) = u(t) + (-1)\cdot v(t)\). This was a crucial step because it allowed the direct application of the Sum Rule. By reorganizing the expression this way, we fit the formula needed to use this helpful rule in differentiation.
Constant Factor Rule
The Constant Factor Rule is another essential tool in calculus, particularly useful when a function is multiplied by a constant. This rule states that when you take the derivative of a function \(f(t)\) that has been multiplied by a constant \(k\), the derivative is simply the constant times the derivative of the function. Formally, this is written as:\[(k \cdot f(t))' = k \cdot f'(t)\]This simplifies the process of taking derivatives, as constants can be "pulled out" and only the function's derivative needs to be calculated.In our exercise, this rule was applied to the expression \((-1) \cdot v(t)\). By using the Constant Factor Rule, we found that the derivative of the term is:\[(d/dt)((-1) \cdot v(t)) = -1 \cdot v'(t)\]This approach helped break down the differentiation step into something more manageable, allowing us to use the result in conjunction with the Sum Rule to reach the final solution.
Function Differentiation
Function differentiation is a foundational concept in calculus. It gives us a systematic way to determine the rate at which one quantity changes in relation to another. Understanding how to differentiate a function involves applying various rules, such as the Sum Rule and the Constant Factor Rule, to figure out how small changes in one variable affect another.Differentiation is represented symbolically by the derivative \(f'(t)\) or using the notation \(d/dt(f(t))\). It tells us the slope of the function at any point, providing insights into the behavior of the function, such as how quickly it's increasing or decreasing.In the context of the exercise, we differentiated the function \(P(t)\). By employing the rules and techniques of differentiation, we were able to transform the composite expression \(P(t) = u(t) - v(t)\) into its instantaneous rate of change, calculated as \(P'(t) = u'(t) - v'(t)\). This illustrates how differentiating a function helps to understand dynamic changes within mathematical relationships.

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