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Chapter 4: Problem 468

Show that \(F(x)\) are antiderivatives of \(f(x)\). \(\quad F(x)=\cos x, f(x)=-\sin x\)

Short Answer

Expert verified
\(F(x) = \cos x\) is an antiderivative of \(f(x) = -\sin x\) because \(F'(x) = -\sin x = f(x)\).

Step by step solution

01

Understand the Definitions

To show that a function \( F(x) \) is an antiderivative of another function \( f(x) \), we need to verify that their derivative relationships satisfy the condition that \( F'(x) = f(x) \).
02

Differentiate \( F(x) \)

Calculate the derivative of \( F(x) = \cos x \). Recall that the derivative of \( \cos x \) with respect to \( x \) is \( -\sin x \). Thus, \( F'(x) = -\sin x \).
03

Compare \( F'(x) \) with \( f(x) \)

After differentiation, we have \( F'(x) = -\sin x \). Now, compare this with the given function \( f(x) = -\sin x \).
04

Verify the Equality

Check that the derivative of \( F(x) \), which is \( F'(x) = -\sin x \), matches exactly with the given function \( f(x) = -\sin x \). Since they are equal, \( F(x) \) is indeed an antiderivative of \( f(x) \).

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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 involves finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point. This is particularly useful for understanding how one variable depends on another.
  • To differentiate a function means to calculate its derivative.
    When you differentiate, you essentially find how the function behaves as you make tiny changes in its variables.
  • In our exercise, we calculated the derivative of \(F(x) = \cos x\).
    By using basic derivative rules, we found that \(F'(x) = -\sin x\).
Understanding differentiation helps solve practical problems in physics, engineering, and economics.
It also lays the groundwork for further study in advanced calculus topics.
Trigonometric Functions
Trigonometric functions are essential in many areas of mathematics and science.
They relate the angles of triangles to the lengths of their sides and provide a way to model periodic phenomena such as sound and light waves.In our given problem, we used two trigonometric functions: cosine and sine.
  • The function \(\cos x\) represents the cosine of an angle, which is the adjacent side over the hypotenuse in a right-angled triangle.
  • The function \(-\sin x\) represents the negative sine of an angle, which is the opposite side over the hypotenuse in a right-angled triangle.
Trigonometric functions have derivatives that follow specific rules.
For example, the derivative of \(\cos x\) is \(-\sin x\), a relationship we used directly in the problem to verify antiderivatives.
Derivative Relationships
Derivative relationships are equations or expressions that show how derivatives of functions are connected to each other.
These relationships help us understand how different functions change and interact as their inputs vary.In the exercise, we confirmed the derivative relationship between \(F(x) = \cos x\) and \(f(x) = -\sin x\).
  • Since the derivative of \(F(x)\), which is \(F'(x)\), matches \(f(x)\), we successfully demonstrated that \(F(x)\) is an antiderivative of \(f(x)\).
  • This means that differentiating \(F(x)\) gives us \(f(x)\), illustrating one of the fundamental aspects of calculus.
Understanding these relationships helps establish foundational links between calculus, geometry, and other mathematical fields.
This allows us to predict function behavior and solve real-world problems using calculus techniques.

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Most popular questions from this chapter

A rocket is launched into space; its kinetic energy is given by \(K(t)=\left(\frac{1}{2}\right) m(t) v(t)^{2},\) where \(K\) is the kinetic energy in joules, \(m\) is the mass of the rocket in kilograms, and \(v\) is the velocity of the rocket in meters/second. Assume the velocity is increasing at a rate of 15 \(\mathrm{m} / \mathrm{sec}^{2}\) and the mass is decreasing at a rate of 10 kg/sec because the fuel is being burned. At what rate is the rocket's kinetic energy changing when the mass is 2000 kg and the velocity is 5000 \(\mathrm{m} / \mathrm{sec}\) ? Give your answer in mega-Joules (MI), which is equivalent to \(10^{6} \mathrm{J} .\)

Determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer. $$ y=\sqrt{4-x^{2}} $$

Find the critical points and the local and absolute extrema of the following functions on the given interval. $$f(x)=3 x^{4}-4 x^{3}-12 x^{2}+6 \text { over }[-3,3]$$

Use a calculator to graph the function over the interval \([a, b]\) and graph the secant line from \(a\) to \(b .\) Use the calculator to estimate all values of \(c\) as guaranteed by the Mean Value Theorem. Then, find the exact value of \(c,\) if possible, or write the final equation and use a calculator to estimate to four digits. $$ [\mathrm{T}] \quad \mathrm{y}=\left|x^{2}+2 x-4\right| \text { over }[-4,0] $$

Use the Mean Value Theorem and find all points \(0 < c < 2\) such that \(f(2)-f(0)=f^{\prime}(c)(2-0)\) $$ f(x)=\sin (\pi x) $$
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