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Chapter 10: Problem 38

Compute the indicated derivative. $$ q(p)=\frac{1}{0.5 p}-3.1 ; q^{\prime}(2) $$

Short Answer

Expert verified
The derivative of the given function \(q(p) = \frac{2}{p} - 3.1\) is \(q'(p) = -2\cdot p^{-2}\). Evaluating at \(p=2\), we get \(q'(2) = -\frac{1}{2}\).

Step by step solution

01

Differentiate \(q(p)\) with respect to \(p\)

We are given the function: \[q(p) = \frac{1}{0.5p} - 3.1\] First, we rewrite the function as: \[q(p) = \frac{2}{p} - 3.1\] Now, we find the derivative of \(q(p)\) with respect to \(p\): \[q'(p) = \frac{d}{dp}\left(\frac{2}{p} - 3.1\right)\] By using the power rule and constant rule, we get: \[q'(p) = -2\cdot p^{-2}\]
02

Evaluate the derivative at \(p=2\)

Now, we plug in \(p=2\) into the derivative expression to find the value of the derivative at this point: \[q'(2) = -2\cdot 2^{-2}\] \[q'(2) = -2\cdot \frac{1}{4}\] \[q'(2) = -\frac{1}{2}\] So, the value of the derivative at \(p=2\) is \(-\frac{1}{2}\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. This process is at the heart of solving problems involving rates of change and can be used to find slopes of tangents to curves and optimize functions, making it a powerful tool in mathematics and the sciences.

For example, to understand the rate at which distance changes with time, we would use differentiation to find the derivative of the distance function with respect to time, which would give us the speed of movement. In the exercise given, differentiation allows us to determine the rate at which the function \( q(p) \) changes when \( p \) changes.
Power Rule
The power rule is a quick way to differentiate functions of the form \( f(x) = x^n \), where \( n \) is any real number. According to the power rule, the derivative of such a function is \( f'(x) = nx^{n-1} \).

In our exercise, when applying the power rule to the term \( \frac{2}{p} \), which is \( 2p^{-1} \) in an exponentiated form, we obtain \( -2p^{-2} \) as the derivative. This simplifies the differentiation process immensely because we only need to multiply the exponent by the coefficient and then subtract one from the exponent to get the derivative.
Constant Rule
In calculus, the constant rule tells us that the derivative of a constant term is zero. This is because a constant does not change, and differentiation measures change. Thus, if there鈥檚 no change, then there is no rate of change to find.

Referring back to the exercise, the term \( -3.1 \) does not depend on \( p \) and therefore its derivative is 0. This rule greatly simplifies calculating derivatives of functions that involve a combination of variable terms and constants, since we can ignore constant terms during differentiation.
Function Notation
Understanding function notation is crucial for successfully studying calculus. Function notation, such as \( f(x) \), clearly communicates that \( f \) is a function with \( x \) as its input variable. It specifies the relationship between the input \( x \) and the output \( f(x) \).

In the context of differentiation, we use primes to denote derivatives; so \( f'(x) \) is the derivative of \( f(x) \) with respect to \( x \). Additionally, we can have multiple variables like in our exercise problem, where we see \( q(p) \) representing a function with \( p \) as the variable, and \( q'(p) \) representing its derivative with respect to \( p \).

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Most popular questions from this chapter

Compute the indicated derivative. $$ L(r)=-1.02 r+5.7 ; L^{\prime}(3.1) $$

Use the method of Example 4 to list approximate values of \(f^{\prime}(x)\) for \(x\) in the given range. Graph \(f(x)\) together with \(f^{\prime}(x)\) for \(x\) in the given range. $$ f(x)=\frac{10 x}{x-2} ; \quad 2.5 \leq x \leq 3 $$

Compute \(f^{\prime}(a)\) algebraically for the given value of a. HINT [See Example 1.] $$ f(x)=2 x-x^{2} ; a=-1 $$

Compute \(f^{\prime}(a)\) algebraically for the given value of a. HINT [See Example 1.] $$ f(x)=-x-x^{2} ; a=0 $$

Use technology to graph the derivative of the given function for the given range of values of \(x .\) Then use your graph to estimate all values of \(x\) (if any) where the tangent line to the graph of the given function is horizontal. Round answers to one decimal place. HINT [See Example 4.] $$ f(x)=-x^{3}-3 x^{2}-1 ; \quad-3 \leq x \leq 1 $$
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