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In Exercises 3–24, use the rules of differentiation to find the derivative of the function. $$ y=\frac{1}{x}-3 \sin x $$

Short Answer

Expert verified
The derivative of the function \(y = \frac{1}{x}-3\sin(x)\) is \(-1x^{-2} - 3\cos(x)\).

Step by step solution

01

Differentiation of the first part

Using the power rule, \(\frac{1}{x}\) can be rewritten as \(x^{-1}\). Differentiating \(x^{-1}\) gives -1 \(x^{-2}\) as derivative.
02

Differentiation of the second part

Using the trigonometric differentiation rule, the derivative of \(\sin(x)\) is \(\cos(x)\). Therefore, the derivative of \(-3\sin(x)\) is -3\(\cos(x)\).
03

Combined derivative

Combine both derivatives to get the derivative of the whole function. The step-by-step derivative of the function \(y = \frac{1}{x}-3\sin(x)\) is thus \(-1x^{-2} - 3\cos(x)\).

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Key Concepts

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

Power Rule
The power rule is a fundamental tool in calculus for differentiating functions that contain variables raised to powers. When you encounter a function with an exponent, the power rule simplifies the process of finding its derivative. Imagine you have a function in the form \(y = x^n\), where \(n\) is a constant exponent. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). This rule is particularly useful because it turns complex expressions into simpler ones with straightforward steps.

In the original exercise, we have \(\frac{1}{x}\), which can be rewritten using exponents as \(x^{-1}\). By applying the power rule here, we differentiate \(x^{-1}\) to get \(-1x^{-2}\).
  • Rewriting the function: \(\frac{1}{x} = x^{-1}\)
  • Applying the power rule: Derivative \(-1x^{-2}\)
Understanding the power rule can transform how you handle derivatives, making it easier to compute derivatives quickly and accurately.
Trigonometric Differentiation
Trigonometric differentiation involves deriving functions that have trigonometric expressions, such as sine, cosine, and tangent. Each trigonometric function has its own specific derivative. For example, the derivative of \(sin(x)\) is \(cos(x)\). This reliable pattern allows for predictable adjustments of trigonometric expressions during differentiation.

In our original problem, we encounter \(-3 \sin(x)\). By applying the trigonometric rule, we know the derivative of \(\sin(x)\) is \(\cos(x)\). Therefore, when differentiating \(-3 \sin(x)\), the constant multiplies the derivative, leading to \(-3 \cos(x)\).
  • Recognize the trigonometric function: \(\sin(x)\)
  • Apply the derivative rule: Derivative \(\cos(x)\)
  • Adjust with constants: \(-3 \sin(x) \) becomes \(-3 \cos(x)\)
By familiarizing yourself with these rules, trigonometric differentiation becomes a simple task.
Derivative of Sin Function
The derivative of the sine function is one of the cornerstones of trigonometric calculus. This function revolves continuously, and its rate of change is captured by its derivative, the cosine function. Understanding this transformation will enlighten you on how periodic functions behave regarding changes in their rates.

Let's focus on the derivative of \(\sin(x)\). The smooth waves of sine transform into the equally smooth waves of cosine when differentiated. This conversion helps handle real-world periodic phenomena, such as sound waves and light patterns.
  • Original function: \(\sin(x)\)
  • Derivative result: \(\cos(x)\)
The trigonometric rules show that the derivative of sinusoidal functions is intuitive once you recognize the patterns between sine and cosine. This fundamental knowledge simplifies various applied calculus problems involving oscillatory motion or wave analysis.

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Most popular questions from this chapter

True or False? In Exercises \(93-96\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If a function has derivatives from both the right and the left at a point, then it is differentiable at that point.

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. (a) Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when the depth \(h\) is 1 foot? (b) The water is rising at a rate of \(\frac{3}{8}\) inch per minute when \(h=2 .\) Determine the rate at which water is being pumped into the trough.

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are $$\begin{array}{l}{P_{1}(x)=f^{\prime}(a)(x-a)+f(a) \text { and }} \\\ {P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)}\end{array}$$ In Exercises 123 and \(124,\) (a) find the specified linear and quadratic approximations of \(f,(b)\) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\) . $$ f(x)=\tan x ; \quad a=\frac{\pi}{4} $$

Using Relationships In Exercises \(103-106,\) use the given information to find \(f^{\prime}(2) .\) $$ \begin{array}{l}{g(2)=3 \quad \text { and } \quad g^{\prime}(2)=-2} \\\ {h(2)=-1 \quad \text { and } \quad h^{\prime}(2)=4}\end{array} $$ $$ f(x)=4-h(x) $$

Acceleration The velocity of an object in meters per second is $$ v(t)=36-t^{2} $$ for \(0 \leq t \leq 6 .\) Find the velocity and acceleration of the object when \(t=3 .\) What can be said about the speed of the object when the velocity and acceleration have opposite signs?

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