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Finding a Derivative of a Trigonometric Function. In Exercises \(39-54,\) find the derivative of the trigonometric function. $$ f(x)=\frac{\sin x}{x^{3}} $$

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{\sin x}{x^{3}}\) is \((\cos x - 3\sin x / x) / x^{2}\)

Step by step solution

01

Identify the numerator and the denominator

In the function \(f(x)=\frac{\sin x}{x^{3}}\), \(\sin x\) is the numerator which we will denote as \(u\), and \(x^{3}\) is the denominator which we will denote as \(v\).
02

Find the derivatives of \(u\) and \(v\)

The derivative of \(\sin x\) is \(\cos x\), so \(u' = \cos x\). The derivative of \(x^{3}\) is \(3x^{2}\), so \(v' = 3x^{2}\).
03

Apply the quotient rule

The quotient rule is \((u/v)' = (vu' - uv') / v^{2}\). Replacing \(u, v, u', v'\) with their respective expressions, we have \((\sin x / x^{3})' = (x^{3}\cos x - \sin x * 3x^{2}) / (x^{3})^{2} = (\cos x - 3\sin x / x) / x^{2}\).

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Key Concepts

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

Quotient Rule
Derivatives of ratios of functions, like \( \frac{\sin x}{x^3} \), require a special technique called the quotient rule. This rule is essential when differentiating a function that is the division of two other functions. The quotient rule states that for two functions \( u \) and \( v \), the derivative \((\frac{u}{v})' \) is calculated as:
  • \((vu' - uv') / v^{2}\)
Think of it as the numerator's derivative multiplied by the denominator, minus the numerator multiplied by the derivative of the denominator. This difference is then divided by the square of the denominator.
It's important because it systematically deals with both parts of the fraction, ensuring the process adheres to the rules of calculus.
In our exercise, using the quotient rule allows us to accurately compute the derivative of \( f(x) = \frac{\sin x}{x^3} \) by blending the derivatives of both \( \sin x \) and \( x^3 \) as shown in the solution.
Chain Rule
While the chain rule doesn't directly apply to our specific problem, it's crucial to understand when dealing with more complex functions, especially those involving compositions of functions.
  • The chain rule is used to find the derivative of a composite function: if you have \( g(f(x)) \), the derivative is \( g'(f(x)) \cdot f'(x) \).
When functions "nest" inside one another, like \( \sin(x^2) \) or \( \cos(3x) \), you'll often need the chain rule. It helps break down operations bit by bit.
In essence, always differentiate the outer function, and then multiply it by the derivative of the inner function. This helps dissect the intricacies of layered equations, making it simpler to handle detailed calculus exercises.
Trigonometric Derivatives
Derivatives of trigonometric functions such as \( \sin, \cos, \) and \( \tan \) are foundational knowledge in calculus. Understanding these derivatives empowers you to tackle a wide array of trigonometric functions.
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \tan x \) is \( \sec^2 x \).
These derivatives stem from the unit circle and limit definitions and form the base of solving more complex problems.
In our exercise, knowing that \( \sin x' = \cos x \) is crucial for applying the quotient rule effectively. Recognizing these patterns makes solving calculus exercises faster and more intuitive.
Calculus Exercises
Solving calculus exercises requires applying various rules and concepts precisely and methodically. With functions involving trigonometric expressions, the approach often combines several fundamental calculus rules.
  • Understand the structure of the function: identify different components like numerators and denominators.
  • Determine which rules apply: quotient rule, chain rule, product rule, etc.
  • Break down the function into manageable parts: compute necessary derivatives step by step.
In our exercise of differentiating \( \frac{\sin x}{x^3} \), we focus on systematically applying known techniques to reach a solution step by step. Practice with diverse problems strengthens your calculus skills, demonstrating how different rules are interwoven in real-world applications.

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Most popular questions from this chapter

Roadway Design Cars on a certain roadway travel on a circular arc of radius \(r .\) In order not to rely on friction alone to overcome the centrifugal force, the road is banked at an angle of magnitude \(\theta\) from the horizontal (see figure). The banking angle must satisfy the equation \(r g \tan \theta=v^{2},\) where \(v\) is the velocity of the cars and \(g=32\) feet per second per second is the acceleration due to gravity. Find the relationship between the related rates \(d v / d t\) and \(d \theta / d t\) .

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\frac{1}{1+x^{2}} ; \frac{d x}{d t}=6 \text { inches per second }} \\ {\begin{array}{ll}{\text { (a) } x=-2} & {\text { (b) } x=0} & {\text { (c) } x=2}\end{array}}\end{array} $$

Using Related Rates In Exercises \(1-4,\) assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t .\) $$ \begin{array}{rlrl}{x y=4} & {\text { (a) } \frac{d y}{d t} \text { when } x=8} & {\frac{d x}{d t}=10} \\ {} & {\text { (b) } \frac{d x}{d t} \text { when } x=1} & {} & {\frac{d y}{d t}=-6}\end{array} $$

Horizontal Tangent Line In Exercises \(73-76\) , determine the point(s) at which the graph of the function has a horizontal tangent line. $$ f(x)=\frac{x-4}{x^{2}-7} $$

Think About It Describe the relationship between the rate of change of \(y\) and the rate of change of \(x\) in each expression. Assume all variables and derivatives are positive. \(\begin{array}{llll}{\text { (a) } \frac{d y}{d t}=3 \frac{d x}{d t}} & {\text { (b) } \frac{d y}{d t}=x(L-x) \frac{d x}{d t},} & {0 \leq x \leq L}\end{array}\)

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