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Chapter 2: Problem 20

Find \(f^{\prime}(x)\). \(f(x)=\sqrt{5} \sin x\)

Short Answer

Expert verified
The derivative is \(f^{\prime}(x)=\sqrt{5} \cos x\).

Step by step solution

01

Identify the function

The given function is \(f(x) = \sqrt{5} \sin x\).
02

Recall the derivative of sine function

Recall that the derivative of \(\sin x\) with respect to \(x\) is \(\cos x\).
03

Apply the constant multiple rule

The constant multiple rule states that the derivative of \(c \cdot g(x)\) is \(c \cdot g^{\prime}(x)\). Here, the constant is \(\sqrt{5}\) and the function \(g(x) = \sin x\).
04

Compute the derivative

Using the constant multiple rule and the derivative of \(\sin x\), we get: \(f^{\prime}(x) = \sqrt{5} \cdot \cos x\).

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Key Concepts

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

Derivative of Trigonometric Functions
Understanding the derivative of trigonometric functions is crucial in calculus. For the function given in the exercise, we particularly focus on \(\sin x\). When taking derivatives, each trigonometric function transforms into another function. For instance, the derivative of \(\sin x\) with respect to \(\ x \) is \(\cos x\). This is a foundational concept to remember:

\(\ \frac{d}{dx} [\sin x] = \cos x \).

Likewise, other common trigonometric derivatives include:
  • \(\ \frac{d}{dx} [\cos x] = -\sin x \)
  • \(\ \frac{d}{dx} [\tan x] = \sec^2 x \)
  • \(\ \frac{d}{dx} [\csc x] = -\csc x \cot x \)
  • \(\ \frac{d}{dx} [\sec x] = \sec x \tan x \)
  • \(\ \frac{d}{dx} [\cot x] = -\csc^2 x \)
Knowing these basic trigonometric derivatives can help simplify many calculus problems.
Constant Multiple Rule
The constant multiple rule is a useful technique in calculus. It states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. Here's the rule:

\[\frac{d}{dx} \big[c \times g(x)\big] = c \times g^{\prime}(x)\]

In our exercise, the constant is \(\sqrt{5}\) and the function is \(\sin x\). Applying the rule, we refer back to the derivative of the \(\sin x\) which is \(\cos x\):

\(\frac{d}{dx} \big[\sqrt{5} \sin x\big] = \sqrt{5} \times \cos x\)

This rule simplifies the differentiation process, making it easier to handle problems involving constants and functions.
Calculus Techniques
Calculus involves various techniques to solve problems involving rates of change and areas under curves. Some of the primary techniques include:
  • Product Rule
  • Quotient Rule
  • Chain Rule
  • Implicit Differentiation
  • Integration
  • Application of Limits
  • First and Second Derivative Tests
In our exercise, we specifically used the constant multiple rule. This technique is a part of the larger family of differentiation rules which include:
  • Understanding basic derivatives like \(\sin x\)
  • Applying rules like the product or chain rule when multiple functions are involved
It's essential to get comfortable with these basic techniques as they form the foundation for solving more complex calculus problems. With practice, these techniques will become intuitive and highly efficient in tackling different types of calculus exercises.

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Most popular questions from this chapter

Differentiate. $$ y=\sqrt{\left(x^{5}+x+1\right)^{3}+7 \sec ^{2} x} $$

Find the derivative of each of the following functions analytically. Then use a grapher to check the results. $$ f(x)=\frac{4 x}{\sqrt{x-10}} $$

Differentiate. $$ y=\sqrt{2+\cos ^{2} t} $$

Differentiate. $$ g(t)=\frac{t^{3}-1}{t^{3}+1} \csc t $$

Differentiate. $$ \sqrt[7]{\csc ^{3}\left(\frac{5 \pi}{6}+2.389\right)} $$
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