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Chapter 5: Problem 1

Compute \(P^{\prime}(t)\) for: a. \(P(t)=e^{5 t}\) b. \(P(t)=\ln 5 t\) c. \(P(t)=e^{t \sqrt{t}}\) d. \(\quad P(t)=e^{\sqrt{2 t}}\) e. \(P(t)=\ln (\ln t)\) f. \(P(t)=e^{\ln t}\) g. \(P(t)=1 /\left(1+e^{t}\right)\) h. \(P(t)=1 / \ln t\) i. \(\quad P(t)=1 /\left(1+e^{-t}\right)\) j. \(\quad P(t)=\left(1+e^{t}\right)^{3}\) k. \(P(t)=\left(e^{\sqrt{t}}\right)^{3} \quad\) l. \(\quad P(t)=\ln \sqrt{t}\)

Short Answer

Expert verified
a. \(5e^{5t}\), b. \(\frac{1}{t}\), c. \(\frac{3\sqrt{t}}{2}e^{t\sqrt{t}}\), d. \(\frac{1}{\sqrt{2t}}e^{\sqrt{2t}}\), e. \(\frac{1}{t\ln(t)}\), f. \(1\), g. \(\frac{-e^{t}}{(1+e^{t})^2}\), h. \(-\frac{1}{t(\ln(t))^2}\), i. \(\frac{e^{-t}}{(1+e^{-t})^2}\), j. \(3e^{t}(1+e^{t})^2\), k. \(\frac{3}{2\sqrt{t}}e^{3\sqrt{t}}\), l. \(\frac{1}{2t}\).

Step by step solution

01

Derive P'(t) for P(t)=e^{5t}

We differentiate the function using the rule \( \frac{d}{dt} e^{ax} = ae^{ax} \). Applying it here, we get:\[ P'(t) = \frac{d}{dt} e^{5t} = 5e^{5t} \]
02

Derive P'(t) for P(t)=\ln(5t)

Use the chain rule where \( \frac{d}{dt} \ln(u) = \frac{1}{u} \cdot \frac{du}{dt} \). Here, \( u = 5t \) and \( \frac{du}{dt} = 5 \):\[ P'(t) = \frac{1}{5t} \cdot 5 = \frac{1}{t} \]
03

Derive P'(t) for P(t)=e^{t\sqrt{t}}

Set \( u = t\sqrt{t} \), then use the chain rule and product rule:\[ \frac{d}{dt}(t\sqrt{t}) = \frac{d}{dt}(t \cdot t^{1/2}) = \sqrt{t} + \frac{t}{2\sqrt{t}} = \frac{3\sqrt{t}}{2} \]Thus:\[ P'(t) = e^{t\sqrt{t}} \cdot \frac{3\sqrt{t}}{2} \]
04

Derive P'(t) for P(t)=e^{\sqrt{2t}}

Set \( u = \sqrt{2t} \), then \( \frac{d}{dt}(\sqrt{2t}) = \frac{1}{2\sqrt{2t}} \cdot 2 = \frac{1}{\sqrt{2t}} \):\[ P'(t) = e^{\sqrt{2t}} \cdot \frac{1}{\sqrt{2t}} \]
05

Derive P'(t) for P(t)=\ln(\ln(t))

Using the chain rule:\[ \frac{d}{dt}(\ln(\ln(t))) = \frac{1}{\ln(t)} \cdot \frac{1}{t} = \frac{1}{t\ln(t)} \]
06

Derive P'(t) for P(t)=e^{\ln(t)}

Simplify \( e^{\ln(t)} = t \). Thus, the derivative is simply:\[ P'(t) = \frac{d}{dt}(t) = 1 \]
07

Derive P'(t) for P(t)=\frac{1}{1+e^{t}}

Use the quotient rule \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \):Here \( u = 1 \), \( du/dt = 0 \), \( v = 1+e^{t} \), \( dv/dt = e^{t} \):\[ P'(t) = \frac{(1+e^{t})\cdot 0 - 1\cdot e^{t}}{(1+e^{t})^2} = \frac{-e^{t}}{(1+e^{t})^2} \]
08

Derive P'(t) for P(t)=\frac{1}{\ln(t)}

Use the quotient rule. Here, \( u = 1 \), \( du/dt = 0 \), \( v = \ln(t) \), \( dv/dt = \frac{1}{t} \):\[ P'(t) = \frac{\ln(t)\cdot 0 - 1\cdot \frac{1}{t}}{(\ln(t))^2} = -\frac{1}{t(\ln(t))^2} \]
09

Derive P'(t) for P(t)=\frac{1}{1+e^{-t}}

Use the quotient rule. Here \( u = 1 \), \( du/dt = 0 \), \( v = 1+e^{-t} \), \( dv/dt = -e^{-t} \):\[ P'(t) = \frac{(1+e^{-t})\cdot 0 - 1\cdot (-e^{-t})}{(1+e^{-t})^2} = \frac{e^{-t}}{(1+e^{-t})^2} \]
10

Derive P'(t) for P(t)=(1+e^{t})^{3}

Using the chain rule \( \frac{d}{dt}(u^n) = nu^{n-1}\frac{du}{dt} \), set \( u = 1+e^{t} \), \( du/dt = e^{t} \):\[ P'(t) = 3(1+e^{t})^{2} \cdot e^{t} \]
11

Derive P'(t) for P(t)=(e^{\sqrt{t}})^{3}

Simplify first to \( e^{3\sqrt{t}} \). Use chain rule with \( u = 3\sqrt{t} \), \( \frac{d}{dt}(3\sqrt{t}) = \frac{3}{2\sqrt{t}} \):\[ P'(t) = e^{3\sqrt{t}} \cdot \frac{3}{2\sqrt{t}} \]
12

Derive P'(t) for P(t)=\ln(\sqrt{t})

Use the chain rule with \( P(t)=\frac{1}{2}\ln(t) \):\[ P'(t) = \frac{1}{2} \cdot \frac{1}{t} = \frac{1}{2t} \]

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Key Concepts

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

Exponential Functions
Exponential functions are one of the most important types of functions in mathematics, characterized by the constant rate of growth or decay. A basic exponential function can be expressed as \( f(x) = a \, e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. The function \( P(t)=e^{5t} \) is a simple form of an exponential function.
To differentiate an exponential function of the form \( e^{ax} \), we use the rule:
  • \( \frac{d}{dt} e^{ax} = ae^{ax} \)
In this context, the growth rate \( a \) is 5, meaning that the derivative is \( 5e^{5t} \). This shows how exponential functions have derivatives that are also exponential functions, making them unique in their simplicity and consistency.
Logarithmic Functions
Logarithmic functions are the inverses of exponential functions. They are useful for solving equations involving exponential growth or decay, like those found in many scientific fields. The natural logarithm, denoted by \( \ln(x) \), is the most common logarithm used in calculus.
To differentiate a natural logarithmic function, the chain rule often comes into play:
  • \( \frac{d}{dt} \ln(u) = \frac{1}{u} \cdot \frac{du}{dt} \)
For instance, if \( P(t)=\ln(5t) \), we identify \( u = 5t \). Thus, the derivative becomes \( \frac{1}{5t}\cdot 5 = \frac{1}{t} \). Logarithmic differentiation reduces complex multiplicative relationships into simpler additive forms, facilitating easier manipulation in calculus.
Chain Rule
The chain rule is a vital tool in calculus that allows us to differentiate composite functions. It is especially useful when the function can be seen as a function within another function. For a composite function \( f(g(t)) \), the chain rule states:
  • \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \)
Consider \( P(t)=e^{t\sqrt{t}} \). Here, the inner function is \( u = t\sqrt{t} \). The derivative of the inner function \( u \) with respect to \( t \) is calculated, and the result is multiplied by the derivative of the outer function \( e^u \). This rule simplifies the differentiation of complex expressions by breaking them into manageable parts.
Product Rule
The product rule is crucial when differentiating functions that are products of two or more functions. If you have a function \( h(t) = f(t) \cdot g(t) \), the product rule is stated as:
  • \( \frac{d}{dt}[f(t)g(t)] = f(t)g'(t) + g(t)f'(t) \)
This rule ensures that we separately consider the rate of change of each component function and how they affect the product. For example, when differentiating \( P(t) = t \cdot t^{1/2} \) (which is part of calculating \( e^{t\sqrt{t}} \)), we differentiate \( t \) and \( t^{1/2} \) separately, then apply the product rule to find the derivative of the product, ensuring an accurate result.

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Most popular questions from this chapter

Suppose penicillin concentration in the serum of a patient \(t\) minutes after a bolus injection of \(2 \mathrm{~g}\) is given by $$P(t)=200 \times 0.96^{t} \quad \mu \mathrm{g} / \mathrm{ml}$$ a. Approximate \(P(t)\) and \(P^{\prime}(t)\) for \(t=0,5,10,15,\) and 20 minutes. b. Plot a graph of \(P^{\prime}(t)\) vs \(P(t)\) using the five pairs of values you just computed.

Find (approximately) equations of the lines tangent to the graphs of a. \(y=1.5^{t}\) at the points \((-1,2 / 3), \quad(0,1),\) and \((1,3 / 2)\) b. \(y=2^{t} \quad\) at the point \(\quad(-1,1 / 2), \quad(0,1),\) and (1,2) c. \(y=3^{t} \quad\) at the point \(\quad(-1,1 / 3), \quad(0,1),\) and (1,3) d. \(y=5^{t} \quad\) at the point \(\quad(-1,1 / 5), \quad(0,1), \quad\) and \(\quad(1,5)\)

A pristine lake of volume \(1,000,000 \mathrm{~m}^{3}\) has a river flowing through it at a rate of \(10,000 \mathrm{~m}^{3}\) per day. A city built beside the river begins dumping \(1000 \mathrm{~kg}\) of solid waste into the river per day. 1\. Write a derivative equation that describes the amount of solid waste in the lake \(t\) days after dumping begins. 2\. What will be the concentration of solid waste in the lake after one year?

We introduced the power chain rule \(\left[(u(x))^{n}\right]^{\prime}=n(u(x))^{n-1}[u(x)]^{\prime}\) for fractional and negative exponents, \(n\), in Section 4.3 .1 (see Exercises 4.3 .3 and 4.3 .4 ). Use these rules when necessary in the following exercise. Compute \(y^{\prime}(x)\) and \(y^{\prime \prime}(x)\) for a. \(y(x)=x^{2}+e^{x}\) b. \(y(x)=3 x^{2}+2 e^{x}\) c. \(y(x)=\left(1+e^{x}\right)^{2}\) d. \(y(x)=\left(e^{x}\right)^{2}\) e. \(y(x)=e^{2 x}=\left(e^{x}\right)^{2}\) f. \(y(x)=e^{-x} \quad=\left(e^{x}\right)^{-1}\) g. \(y(x)=e^{3 x} \quad=\left(e^{x}\right)^{3}\) h. \(y(x)=e^{x} \times e^{2 x}\) i. \(y(x)=\left(5+e^{x}\right)^{3}\) j. \(y(x)=\frac{1}{1+e^{x}} \quad=\left(1+e^{x}\right)^{-1}\) k. \(y(x)=\sqrt{e^{x}} \quad=\left(e^{x}\right)^{\frac{1}{2}}\) l. \(y(x)=e^{\frac{1}{2} x}\) m. \(y(x)=e^{0.6 x}=\left(e^{x}\right)^{0.6}\) n. \(y(x)=e^{-0.005 x}\)

Find values of \(C\) and \(k\) so that \(P(t)=C e^{k t}\) matches the data. a. \(P(0)=5 \quad P(2)=10\) b. \(P(0)=10 \quad P(2)=5\) c. \(P(0)=2 \quad P(5)=10\) d. \(P(0)=10 \quad P(5)=10\) e. \(P(0)=5 \quad P(2)=2\) f. \(P(0)=8 \quad P(10)=6\) g. \(P(1)=5 \quad P(2)=10\) h. \(P(2)=10 \quad P(10)=20\)
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