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Chapter 7: Problem 36

\(F(s)=\arctan (1 / s)\)

Short Answer

Expert verified
The derivative of the given function is \( F'(s) = - \frac{1}{s^2 + 1} \)

Step by step solution

01

Identify the Function

The given function is \(F(s)=\arctan (1 / s)\).
02

Derivative of the Arctangent

Recall that the derivative of arctan(u) is \( \frac{1}{1 + u^2} \cdot u'\) where \(u'\) is the derivative of \(u\).
03

Compute Derivative

Differentiate \(F(s)\) with respect to \(s\), using the chain rule (the derivative of the outside function times the derivative of the inside function). Here, \(u = 1 / s\). So \(u' = - 1 / s^2\). Plugging into the formula above we find the derivative of the function is \(F'(s) = \frac{1}{1 + (1 / s)^2} \cdot (- 1 / s^2)\). Simplifying this gives us \(F'(s) = - \frac{1}{s^2 + 1}\)

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Key Concepts

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

Understanding the Derivative of Arctangent
The arctangent function, denoted as \(\arctan(x)\), is the inverse of the tangent function. Understanding its derivative is crucial in calculus. The derivative of \(\arctan(u)\) with respect to \(u\) is given by the formula:
\[ \frac{1}{1 + u^2} \cdot u' \] Here, \(u'\) refers to the derivative of the inner function \(u\) concerning the variable you are differentiating with respect to.
  • The expression \(1 + u^2\) is vital as it accounts for how the function changes over its domain.
  • This formula reveals that the derivative decreases as \(|u|\) increases, making \(\arctan\) a smooth and continuous function.
To find the derivative of a function like \(F(s)=\arctan(1/s)\), identify \(u\) as \(1/s\), then use this formula to differentiate.
Applying the Chain Rule
The chain rule is a fundamental principle in calculus that helps us differentiate composite functions. When a function is nested inside another, the chain rule allows us to systematically break it down.
The function \(F(s) = \arctan(1/s)\) requires the chain rule because it involves a function within a function.
  • The outer function here is \(\arctan\), and the inner function is \(1/s\).
  • The chain rule states: to differentiate these, you multiply the derivative of the outer function by the derivative of the inner function.
First, determine the derivative of \(\arctan(u)\), and then multiply it by the derivative of \(1/s\), which is \(-1/s^2\).
This process effectively links the two functions together, revealing the overall derivative of \(F(s)\).
Basics of Function Differentiation
Differentiation is the process of finding a derivative, which is essentially the rate at which a function changes. It is a cornerstone concept in calculus. For any function \(f(x)\), the derivative \(f'(x)\) provides critical information:
  • It tells us how the output value changes as the input value changes.
  • This information is crucial for understanding curves, slopes, and tangents.
In practice, differentiating \(F(s) = \arctan(1/s)\) means applying basic rules carefully:
  • Recognize the function structure; here, it blends inverse trigonometry and algebra.
  • Use differentiation techniques, such as the chain rule, to systematically find derivatives.
Once the derivative is calculated, it can be simplified to provide insights into the rate and direction of change at any \(s\) point in the domain.

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Most popular questions from this chapter

Verify \((3)\) in Theorem 9 for the function \(f(t)=\sin t\) taking the period as 2\(\pi .\) Repeat, taking the period as 4\(\pi .\)

\(\begin{array}{ll}{x^{\prime}=x-y ;} & {x(0)=-1} \\ {y^{\prime}=2 x+4 y ;} &{y(0)=0}\end{array}\)

\(g(t)=\left\\{\begin{array}{ll}{0,} & {0< t <2} \\ {t+1,} & {2< t}\end{array}\right.\)

Thanks to Euler's formula (page 166) and the algebraic properties of complex numbers, several of the entries of Table 7.1 can be derived from a single formula; namely, (6) $$\quad \mathscr{L}\left\\{e^{(a+i b) t}\right\\}(s)=\frac{s-a+i b}{(s-a)^{2}+b^{2}}, \quad s>a$$ (a) By computing the integral in the definition of theLaplace transform on page 353 with \(f(t)=e^{(a+i b) t}\)show that $$\mathscr{L}\left\\{e^{(a+i b) t}\right\\}(s)=\frac{1}{s-(a+i b)}, \quad s>a$$ (b) Deduce (6) from part (a) by showing that$$\frac{1}{s-(a+i b)}=\frac{s-a+i b}{(s-a)^{2}+b^{2}}$$ (c) By equating the real and imaginary parts in formula (6), deduce the last two entries in Table 7.1.

\((D-4)[x]+6 y=9 e^{-3 r} ; \quad x(0)=-9$$x-(D-1)[y]=5 e^{-3 t} ; \quad y(0)=4\)
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