/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Calculate the derivative of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the derivative of the following functions. $$y=\sin ^{5}(\cos 3 x)$$

Short Answer

Expert verified
Answer: The derivative of the function $$y = \sin^5(\cos 3x)$$ is $$\frac{dy}{dx} = -15(\sin^4 (\cos 3x))\cdot \cos(\cos 3x) \cdot \sin(3x)$$.

Step by step solution

01

Identify the outer and inner functions

The given function can be thought of as a composition of three functions: 1. The outer function: $$f(u) = u^5$$ 2. The middle function: $$g(x) = \sin(v)$$ 3. The inner function: $$h(x) = \cos(3x)$$ In other words, $$y = f(g(h(x)))$$.
02

Differentiate the outer, middle, and inner functions

Now we will find the derivatives of these functions f(u), g(v), and h(x) with respect to their respective variables: 1. Derivative of the outer function: $$f'(u) = 5u^4$$ 2. Derivative of the middle function: $$g'(v) = \cos(v)$$ 3. Derivative of the inner function: $$h'(x) = -3\sin(3x)$$
03

Use the chain rule to find the derivative of y

According to the chain rule, the derivative of the composite function y is the product of the derivatives of the outer, middle, and inner functions: $$\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$ Now substitute the values of their respective derivatives.
04

Substitute and simplify

Substitute the derivatives and simplify the expression: $$\frac{dy}{dx} = 5(\sin (\cos 3x))^4\cdot \cos(\cos 3x) \cdot (-3\sin 3x)$$ Finally, we arrive at the derivative of the given function: $$\frac{dy}{dx} = -15(\sin^4 (\cos 3x))\cdot \cos(\cos 3x) \cdot \sin(3x)$$ The derivative of the function $$y=\sin ^{5}(\cos 3 x)$$ is $$\frac{dy}{dx} = -15(\sin^4 (\cos 3x))\cdot \cos(\cos 3x) \cdot \sin(3x)$$.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

The Chain Rule
The chain rule is an essential tool in calculus, especially when dealing with composite functions. It allows us to differentiate functions that are composed of two or more simpler functions. The basic idea of the chain rule is to find the derivative of the outer function and multiply it by the derivative of the inner function.
For example, if you have a function that can be expressed as \( y = f(g(x)) \), the chain rule states that the derivative, \( \frac{dy}{dx} \), is given by:
  • First find \( f'(g(x)) \), which is the derivative of the outer function evaluated at the inner function.
  • Next, multiply by \( g'(x) \), the derivative of the inner function.
This results in \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). By using this process, you can effectively break down a complex differentiation problem into manageable parts.
It's especially useful in higher-level mathematics and real-world applications where functions are rarely simple.
Trigonometric Functions
Trigonometric functions, like sine and cosine, play a crucial role in calculus and geometry. They are periodic functions that help model phenomena such as sound waves, light, and other oscillatory motions. In our original problem, we encounter the sine and cosine functions.
The sine function, \( \sin(x) \), relates the angle of a right-angled triangle to the ratio of the opposite side over the hypotenuse. Similarly, the cosine function, \( \cos(x) \), relates the angle to the ratio of the adjacent side over the hypotenuse.
  • The derivative of \( \sin(x) \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) is \(-\sin(x) \).
Understanding these derivatives is essential when applying the chain rule to composite functions involving trigonometry. These functions help us describe and predict cyclic behaviors, making them indispensable in physics and engineering.
Composite Functions
Composite functions are functions made by combining two or more other functions. This means that the output of one function becomes the input of another. In the given problem, we have a complex function \( y = \sin^5(\cos(3x)) \), which is a composite of three simpler functions.
We can think of composite functions as layers of an onion, where you peel back each layer (or function) one at a time. Here's how our function breaks down:
  • Outer function: \( f(u) = u^5 \). This wraps around the middle function.
  • Middle function: \( g(v) = \sin(v) \). This applies the sine transformation.
  • Inner function: \( h(x) = \cos(3x) \). This applies the cosine transformation and scales the input.
To differentiate such a function, you must apply the chain rule to each layer. Start with the outermost function and work inward. Composite functions are widespread in both pure and applied mathematics, as they offer a flexible framework for modeling complex relationships.
By breaking functions into their composite parts, you simplify analysis and gain deeper insights into their behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Robert Boyle \((1627-1691)\) found that for a given quantity of gas at a constant temperature, the pressure \(P\) (in kPa) and volume \(V\) of the gas (in \(m^{3}\) ) are accurately approximated by the equation \(V=k / P\), where \(k>0\) is constant. Suppose the volume of an expanding gas is increasing at a rate of \(0.15 \mathrm{m}^{3} / \mathrm{min}\) when the volume \(V=0.5 \mathrm{m}^{3}\) and the pressure is \(P=50 \mathrm{kPa}\). At what rate is pressure changing at this moment?

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At \(40^{\circ}\) north latitude, the length of a day is approximated by $$D(t)=12-3 \cos \left(\frac{2 \pi(t+10)}{365}\right)$$ where \(D\) is measured in hours and \(0 \leq t \leq 365\) is measured in days, with \(t=0\) corresponding to January 1 a. Approximately how much daylight is there on March 1 \((t=59) ?\) b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March \(1 .\) Convert your answer to units of min/day and explain what this result means. d. Graph the function \(y=D^{\prime}(t)\) using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

Two boats leave a port at the same time, one traveling west at \(20 \mathrm{mi} /\) hr and the other traveling southwest \(\left(45^{\circ}\right.\) south of west) at \(15 \mathrm{mi} / \mathrm{hr}\). After 30 minutes, how far apart are the boats and at what rate is the distance between them changing? (Hint: Use the Law of Cosines.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.