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Chapter 27: Problem 34

Find the derivatives of the given functions. $$y=2 x \sin x+2 \cos x-x^{2} \cos x$$

Short Answer

Expert verified
The derivative of the function is: \( y' = 4x \cos x - x^2 \sin x \)

Step by step solution

01

Identify Functions and Their Derivatives

The given function is a combination of products and trigonometric functions. The function is: \[ y = 2x \sin x + 2 \cos x - x^2 \cos x \] We need to use the product rule and standard derivatives for \( \sin x \) and \( \cos x \) which are \( \cos x \) and \( -\sin x \) respectively.
02

Apply Product Rule to Each Term Separately

The function consists of three terms. We'll break them down:- For the first term \( 2x \sin x \), we use the product rule: If \( u = 2x \) and \( v = \sin x \), then \( u' = 2 \) and \( v' = \cos x \). The derivative using product rule is \( uv' + vu' = 2x \cos x + 2 \sin x \).- The second term \( 2\cos x \) is straightforward, the derivative is \( -2\sin x \).- For the third term \( x^2 \cos x \), again use the product rule: If \( u = x^2 \) and \( v = \cos x \), then \( u' = 2x \) and \( v' = -\sin x \). The derivative is \( uv' + vu' = x^2(-\sin x) + 2x\cos x \).
03

Combine Derivatives of Each Term

Now combine the derivatives from each term to form the complete derivative of the function:- From the first term: \( 2x \cos x + 2 \sin x \)- From the second term: \( -2 \sin x \)- From the third term: \( -x^2 \sin x + 2x \cos x \)Combine them:\[ y' = (2x \cos x + 2 \sin x) + (-2 \sin x) + (-x^2 \sin x + 2x \cos x) \]
04

Simplify the Expression

Simplify the derivative expression obtained in Step 3:- Combine like terms:\[ y' = 2x \cos x + 2 \sin x - 2 \sin x - x^2 \sin x + 2x \cos x \]- Notice that \( 2 \sin x \) and \( -2 \sin x \) cancel each other out.- Combining terms gives:\[ y' = 4x \cos x - x^2 \sin x \]

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Key Concepts

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

Understanding Derivatives
Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. Think of derivatives as the way to quantify the rate of change or the slope of a function at any given point.
If you're driving a car and you look at the speedometer, you're seeing a real-world example of a derivative. The speedometer tells you how fast your distance is changing with respect to time.
  • A derivative of a function is often represented by \( y' \) or \( \frac{dy}{dx} \).
  • To find the derivative, we often use rules like the product rule if the function involves multiplication.
  • Trigonometric functions like \( \sin x \) or \( \cos x \) have specific derivatives, namely \( \cos x \) and \( -\sin x \), respectively.
Applying the Product Rule
The product rule is a method used to differentiate functions that are the product of two smaller functions. It helps us find the derivative of complex expressions more easily.
If you have a product of two functions, say \( u(x) \) and \( v(x) \), the product rule states:
  • The derivative of \( u(x)v(x) \) is \( u'v + uv' \).

Here's how it works in practice:
  • Consider the expression \( 2x \sin x \) from our problem.
  • Let \( u = 2x \) and \( v = \sin x \).
  • Find the derivatives: \( u' = 2 \) and \( v' = \cos x \).
  • Apply the product rule: \( u'v + uv' = 2x \cos x + 2 \sin x \).
Using the product rule simplifies the process significantly, particularly when you have multiple terms to deal with.
Basics of Trigonometric Functions
Trigonometric functions like \( \sin x \) and \( \cos x \) frequently appear in calculus problems. Let's delve into their behavior and derivatives:
  • \( \sin x \) is a periodic function that oscillates between -1 and 1. Its graph looks like smooth waves.
  • The derivative of \( \sin x \) is \( \cos x \). This tells us that the rate of change of \( \sin x \) at any point is the value of \( \cos x \) at that same point.
  • \( \cos x \) complements \( \sin x \) and is also periodic, oscillating similarly.
  • The derivative of \( \cos x \) is \( -\sin x \). When the value of \( \cos x \) decreases, \( -\sin x \) goes negative, reflecting the inverse relationship.
Understanding these functions is crucial for mastering calculus as they form the basis for solving many physics and engineering problems.
Simplifying Expressions
Simplifying expressions involves reducing a complex derivative or mathematical expression to its simplest form. This makes it easier to evaluate, compare, or graph.
In the step-by-step solution, after applying the product rule and finding the derivatives for all terms, the next goal was to simplify the expression:
  • You often combine like terms. For example, consider terms like \( 2 \sin x \) and \( -2 \sin x \) which cancel out, simplifying the expression.
  • In our problem, after combining terms, we ended with \( 4x \cos x - x^2 \sin x \).
  • Simplifying helps reveal patterns or symmetry in functions, providing deeper insights into the behavior of the function.
Remember, keeping expressions clean and simple not only helps you, but also makes it clearer when communicating math solutions to others.

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Most popular questions from this chapter

Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{t \rightarrow \infty} \frac{\ln \ln t}{\ln t}$$

Solve the given problems by finding the appropriate derivative. An object on the end of a spring is moving so that its displacement (in \(\mathrm{cm}\) ) from the equilibrium position is given by \(y=e^{-0.5 t}(0.4 \cos 6 t-0.2 \sin 6 t) .\) Find the expression for the velocity of the object. What is the velocity when \(t=0.26 \mathrm{s} ?\) The motion described by this equation is called damped harmonic motion.

Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{t \rightarrow \pi / 4} \frac{1-\sin 2 t}{\frac{\pi}{4}-t}$$

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.30 shows the graph of \(y=\sinh x\). The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{u}+e^{-u}\right)\) Figure 27.31 shows the graph of \(y=\cosh x\). These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\). Show that \(\frac{d}{d x} \sinh u=\cosh u \frac{d u}{d x}\) and \(\frac{d}{d x} \cosh u=\sinh u \frac{d u}{d x}\) where \(u\) is a function of \(x\).

Solve the given problems. Another indeterminate form is \(\infty-\infty\). Often, it is possible to make an algebraic or trigonometric change in the function so that it will take on the form of a \(0 / 0\) or \(\infty / \infty\) indeterminate form. Find \(\lim _{\theta \rightarrow \frac{\pi}{2}}(\sec \theta-\tan \theta)\)
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