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Chapter 15: Problem 2

Find the rate of change of the following functions: (a) \(\mathrm{e}^{t} \pm \mathrm{e}^{-t}\) (d) \(\sqrt{r}+2 r^{2}\) (c) \(-3 \cos x\) (e) \(2 \mathrm{e}^{-0.5 v}+v^{3}\)

Short Answer

Expert verified
(a) \(e^t \mp e^{-t}\), (d) \(\frac{1}{2}r^{-1/2} + 4r\), (c) \(3\sin x\), (e) \(-e^{-0.5v} + 3v^2\)."

Step by step solution

01

Differentiate function (a)

For the function \( e^t \pm e^{-t} \), we need to find its derivative with respect to \( t \). Using the rules of differentiation, the derivative of \( e^t \) is \( e^t \) and the derivative of \( e^{-t} \) is \( -e^{-t} \). Therefore, the rate of change is:\[\frac{d}{dt}(e^t \pm e^{-t}) = e^t \mp e^{-t}.\]
02

Differentiate function (d)

The function is \( \sqrt{r} + 2r^2 \). We start by rewriting \( \sqrt{r} \) as \( r^{1/2} \). The derivative of \( r^{1/2} \) is \( \frac{1}{2}r^{-1/2} \). The derivative of \( 2r^2 \) is \( 4r \). Thus, the rate of change is:\[\frac{d}{dr} \left( \sqrt{r} + 2r^2 \right) = \frac{1}{2}r^{-1/2} + 4r.\]
03

Differentiate function (c)

For the function \( -3 \cos x \), find the derivative with respect to \( x \). The derivative of \( \cos x \) is \( -\sin x \), thus the derivative of \( -3 \cos x \) will be \( -3(-\sin x) = 3 \sin x \). So, the rate of change is:\[\frac{d}{dx}(-3 \cos x) = 3 \sin x.\]
04

Differentiate function (e)

The function is \( 2e^{-0.5v} + v^3 \). For \( 2e^{-0.5v} \), apply the chain rule to find the derivative: \( -0.5 \times 2e^{-0.5v} = -e^{-0.5v} \). For \( v^3 \), the derivative is \( 3v^2 \). Thus, the rate of change is:\[\frac{d}{dv} (2e^{-0.5v} + v^3) = -e^{-0.5v} + 3v^2.\]

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Key Concepts

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

Rates of Change
In mathematics, the **rate of change** is an important concept that tells us how a function's value changes with respect to its input variables. It's essentially the core idea of differentiation. Think of it like how fast a car is going 鈥� it's about the speed or velocity, except here it's about the speed of change of a mathematical function.
  • For instance, if you have a function that represents the position of an object over time, its rate of change (or derivative) would tell you the object's velocity.
  • In our exercises, we are looking at the rate at which the functions given change with respect to specific variables. So, when you differentiate a function, you're finding its rate of change.
To find this rate, we use the process of differentiation. It involves applying rules to calculate how a small change in the input affects the function's output.
Derivative Rules
Finding the derivative of a function often requires using specific **rules of differentiation**. These rules are foundational tools in calculus that simplify the process.
Some important rules include:
  • **Power Rule**: If you have a function of the form \( x^n \), the derivative is \( nx^{n-1} \).
  • **Sum Rule**: When differentiating sums of functions, like \( u(x) + v(x) \), you take the derivative of each function separately: \( u'(x) + v'(x) \).
  • **Exponential Rule**: For functions involving exponential terms, \( e^x \) differentiates to \( e^x \), and if the exponent involves more variables, you'll often apply the chain rule, too.
In our example, using these rules allowed us to find the rate of change for functions involving exponentials and powers of variables.
Chain Rule
The **chain rule** is a valuable differentiation technique used when dealing with composite functions 鈥� functions within functions. It's like peeling an onion; you have to go layer by layer.
Here's how you might apply it:
  • If you have a composite function \( f(g(x)) \), the derivative is found using: \( f'(g(x)) \cdot g'(x) \).
  • Essentially, you take the derivative of the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.
In our exercise, the function \( 2e^{-0.5v} \) required the use of the chain rule. Here, \( -0.5v \) was treated as the inside function, and the exponential the outside function. By applying the chain rule, we found how such composite structures change with respect to their input variable.
Trigonometric Derivatives
**Trigonometric derivatives** are special derivative rules that apply to trigonometric functions like sine, cosine, and tangent.
Some key trigonometric derivatives are:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • These derivatives are essential when dealing with trigonometric functions, as they follow a pattern based on trigonometric identities.
In our step-by-step solution, we saw the derivative of \( -3\cos x \), which utilized the derivative rule for cosine. Understanding these derivatives helps us quickly determine how these trigonometric functions change, a crucial skill when solving calculus problems.

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Most popular questions from this chapter

Explain the meaning of the expression \(\frac{\mathrm{d} y}{\mathrm{~d} x}\).

Calculate \(y^{\prime \prime}\) where \(y\) is given by (a) \(\cos 2 t-\sin 2 t\) (c) \(2 x^{6}-3 x^{7}\) (d) \(-x^{3}+3 x^{2}\) (e) \(9-\frac{9}{x}\)

The function \(y(x)\) is defined by $$ y(x)=x^{4}-3 x^{3}+3 x^{2}+1 $$ Calculate the values of \(x\) where \(y^{\prime \prime}=0\)

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