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Chapter 3: Problem 29

Find the derivatives of the following functions. $$P=\frac{40}{1+2^{-t}}$$

Short Answer

Expert verified
The derivative of the function, \(\frac{dP}{dt}\), is: \(\frac{dP}{dt}=\frac{40\cdot 2^{-t}\ln(2)}{(1+2^{-t})^2}\)

Step by step solution

01

Identify function components

Identify the numerator \(u\) and the denominator \(v\) of the function: $$u=40$$ $$v=1+2^{-t}$$
02

Find derivatives of u and v

Calculate the derivative of the numerator with respect to \(t\): $$u'=\frac{d}{dt}(40)=0$$ Calculate the derivative of the denominator with respect to \(t\): $$v'=\frac{d}{dt}(1+2^{-t})=\frac{d}{dt}(1)+\frac{d}{dt}(2^{-t})=0-2^{-t}\ln(2)=-2^{-t}\ln(2)$$
03

Apply the Quotient Rule

Apply the Quotient Rule to find the derivative of the function using the components we found in Steps 1 and 2: $$\frac{dP}{dt} = \frac{vu' - uv'}{v^2}$$
04

Calculate the derivative

Now substitute the values for \(u\), \(v\), \(u'\), and \(v'\) found earlier and calculate the derivative: $$\frac{dP}{dt} = \frac{40\cdot0 - 40\cdot(-2^{-t}\ln(2))}{(1+2^{-t})^2}=\frac{40\cdot 2^{-t}\ln(2)}{(1+2^{-t})^2}$$ #Answer# Using the steps above, the derivative of the given function is: $$\frac{dP}{dt}=\frac{40\cdot 2^{-t}\ln(2)}{(1+2^{-t})^2}$$

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Key Concepts

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

Quotient Rule
The Quotient Rule is essential when finding the derivative of a function that is a quotient, or fraction, of two other functions. It is especially useful when there is a more complex function divided by another, simpler function. The Quotient Rule states that if you have a function that is the quotient of two functions, say \( rac{u}{v} \), its derivative is given by:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
  • \( u \) is the numerator function.
  • \( v \) is the denominator function.
  • \( u' \) is the derivative of the numerator.
  • \( v' \) is the derivative of the denominator.
Apply this rule by carefully identifying which part is the numerator and which part is the denominator. Then, find their respective derivatives.
Plug these results into the formula above to compute the derivative of the entire fraction. Remember that careful application will ensure accuracy.
exponential functions
Exponential functions involve powers of a constant base raised to a variable exponent. A basic form looks like \( a^x \), where \( a \) is a constant.
In calculus, understanding the differentiation of these is crucial since their structure allows for growth and decay models in real-life scenarios.For the derivative of an exponential function of the form \( a^x \), you use the formula: \[\frac{d}{dx}(a^x) = a^x \ln(a)\]- \( a^x \) remains mostly unchanged after differentiation.- The factor \( \ln(a) \) is multiplied by the remaining function.This unique property makes exponential functions wildly different from polynomial functions.
In our exercise, this is seen when differentiating \( 2^{-t} \), which is modified as \( -2^{-t} \ln(2) \) during derivation. The negative sign emerges from the negative exponent.
chain rule
The chain rule is a vital tool for finding the derivative of composite functions. These are functions within functions, a scenario frequently encountered in calculus problems. If you have a function made up of two functions, such as \( f(g(x)) \), the chain rule helps you find its derivative. The formula is:\[\frac{dy}{dx} = f'(g(x)) \, g'(x)\]Here's how it works:
  • Step 1: Differentiate the outer function \( f \), keeping the inner function \( g(x) \) intact.
  • Step 2: Multiply the result by the derivative of the inner function \( g'(x) \).
The chain rule is not explicitly detailed in every exercise step. However, it underlies many of the principles applied, such as when differentiating composite parts of a function.
Recognizing composite functions let you apply the chain rule efficiently to handle even the most complicated derivatives.

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Most popular questions from this chapter

An airliner passes over an airport at noon traveling \(500 \mathrm{mi} / \mathrm{hr}\) due west. At \(1: 00 \mathrm{p} . \mathrm{m} .,\) another airliner passes over the same airport at the same elevation traveling due north at \(550 \mathrm{mi} / \mathrm{hr} .\) Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 2: 30 p.m.?

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}(f(x) g(x))\right|_{x=1}$$

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(2 \sin ^{-1} x\right)=1-2 x^{2}$$

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}}$$

Electrostatic force The magnitude of the electrostatic force between two point charges \(Q\) and \(q\) of the same sign is given by \(F(x)=\frac{k Q q}{x^{2}},\) where \(x\) is the distance (measured in meters) between the charges and \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\) is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force). a. Find the instantaneous rate of change of the force with respect to the distance between the charges. b. For two identical charges with \(Q=q=1 \mathrm{C},\) what is the instantaneous rate of change of the force at a separation of \(x=0.001 \mathrm{m} ?\) c. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain.
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