/*! 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 37 Find the derivative of the follo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative of the following functions. $$y=x \cos x \sin x$$

Short Answer

Expert verified
Question: Find the derivative of the function $$y = x \cos x \sin x$$. Answer: The derivative of the function is $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$.

Step by step solution

01

Identify the functions and their derivatives

We need to find the derivatives of the three functions: 1. $$u(x) = x$$, so its derivative is $$u'(x) = 1$$ 2. $$v(x) = \cos x$$, so its derivative is $$v'(x) = -\sin x$$ 3. $$w(x) = \sin x$$, so its derivative is $$w'(x) = \cos x$$
02

Apply the product rule for three functions

The product rule for three functions is as follows: $$(uvw)' = u'vw + uv'w + uvw'$$ In our case, the function $$y(x) = uvw$$, and we will substitute the function and its derivatives into the equation. $$y'(x) = (1)(\cos x)(\sin x) + (x)(-\sin x)(\sin x) + (x)(\cos x)(\cos x)$$
03

Simplify the expression

Simplify the expression to get the final derivative. $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$ So, the derivative of the function $$y = x \cos x \sin x$$ is: $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$

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.

Product Rule
The product rule is an essential tool in calculus used for finding the derivative of the product of two or more functions. It's especially useful when you can't simplify a function into its individual components for easier differentiation. For two functions, the product rule states that if you have functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:
  • \((uv)' = u'v + uv'\)
This means that the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second. When dealing with three functions, the formula extends to:
  • \((uvw)' = u'vw + uv'w + uvw'\)
Here, you differentiate each function one by one, keeping the others constant, and then add the results. In our given problem, applying the product rule helps streamline the differentiation process for the function \( y = x \cos x \sin x \). By treating each term according to the rule, you can tackle even the most complicated of expressions.
Trigonometric Functions
Trigonometric functions such as \( \cos x \) and \( \sin x \) are fundamental in mathematics, especially in calculus. These functions relate the angles of a triangle to the lengths of its sides and have various properties used in differentiation and integration. When differentiating trigonometric functions:
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \sin x \) is \( \cos x \).
These results are crucial when applying differentiation techniques to mathematical expressions involving trigonometric functions. Recognizing the derivatives of these functions allows you to quickly apply rules like the product rule and simplify complex expressions. By observing these foundational derivatives, differentiating combinations of trigonometric functions becomes more manageable.
Differentiation Techniques
Differentiation techniques allow you to find the rate at which something changes, and they play a pivotal role in calculus. Beyond the basic rules of differentiation such as the power, constant, and sum rules, more specialized techniques include the product rule, quotient rule, and chain rule. This diversity lets you tackle a wide range of functions and combinations effectively. In this exercise, we used:
  • The product rule to differentiate a product of three functions.
  • Knowledge of basic trigonometric derivatives.
Other common techniques include:
  • The chain rule, used when functions are composed of other functions.
  • The quotient rule, applicable for ratios of functions.
By mastering these techniques, you can approach calculus problems methodically, breaking down each component function, and applying the appropriate rule. This structured approach ensures you handle each piece accurately, leading to a correct overall solution. These methods are crucial tools in your mathematical toolkit, allowing you to solve intricate problems efficiently.

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

The lapse rate is the rate at which the temperature in Earth's atmosphere decreases with altitude. For example, a lapse rate of \(6.5^{\circ}\) Celsius / km means the temperature decreases at a rate of \(6.5^{\circ} \mathrm{C}\) per kilometer of altitude. The lapse rate varies with location and with other variables such as humidity. However, at a given time and location, the lapse rate is often nearly constant in the first 10 kilometers of the atmosphere. A radiosonde (weather balloon) is released from Earth's surface, and its altitude (measured in kilometers above sea level) at various times (measured in hours) is given in the table below. $$\begin{array}{lllllll} \hline \text { Time (hr) } & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \text { Altitude (km) } & 0.5 & 1.2 & 1.7 & 2.1 & 2.5 & 2.9 \\ \hline \end{array}$$ a. Assuming a lapse rate of \(6.5^{\circ} \mathrm{C} / \mathrm{km},\) what is the approximate rate of change of the temperature with respect to time as the balloon rises 1.5 hours into the flight? Specify the units of your result and use a forward difference quotient when estimating the required derivative. b. How does an increase in lapse rate change your answer in part (a)? c. Is it necessary to know the actual temperature to carry out the calculation in part (a)? Explain.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$y=x e^{y}$$

Find \(d^{2} y / d x^{2}.\) $$x+y=\sin y$$

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{2}+x y+y^{2}=7 ;(2,1)$$ (Graph cant copy)

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

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.