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Chapter 2: Problem 7

Find the derivative of \(f(x)=9 x \sin (2 x)\) \(f^{\prime}(x)=\) _________

Short Answer

Expert verified
The derivative of \(f(x) = 9x \, \sin(2x)\) is \(f'(x) = 9 \sin(2x) + 18x \cos(2x)\).

Step by step solution

01

Identify the components of the function

Given function is: \[f(x) = 9x \, \sin(2x)\]This is a product of two functions: \(u(x) = 9x\) and \(v(x) = \sin(2x)\).
02

Apply the product rule

To find the derivative of \(f(x) = 9x \, \sin(2x)\), the product rule states: \[f'(x) = u'(x)v(x) + u(x)v'(x)\]Now, compute \(u'(x)\) and \(v'(x)\).
03

Differentiate \(u(x)\)

Differentiating \(u(x) = 9x\): \[u'(x) = 9\]
04

Differentiate \(v(x)\) using the chain rule

Given \(v(x) = \sin(2x)\), apply the chain rule. Let \(g(x) = 2x\) and \(h(g) = \sin(g)\): \[v'(x) = \frac{d}{dx} \sin(2x) = \cos(2x) \, \frac{d}{dx}(2x) = \cos(2x) \, 2 = 2 \cos(2x)\]
05

Combine the results using the product rule

Now, substitute \(u'(x)\), \(u(x)\), \(v(x)\), and \(v'(x)\) back into the product rule: \[f'(x) = 9 \, \sin(2x) + 9x \, 2\cos(2x)\]
06

Simplify the expression

Simplify the derivative expression: \[f'(x) = 9 \sin(2x) + 18x \cos(2x)\]

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Key Concepts

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

The Product Rule
The product rule is essential when you're working with the derivative of a product of two functions. It's stated as follows: If you have two functions, \(u(x)\) and \(v(x)\), and you want to differentiate their product \(f(x) = u(x)v(x)\), the product rule tells you to use the formula:

\[f'(x) = u'(x)v(x) + u(x)v'(x)\]

This means you first take the derivative of \(u(x)\), multiply it by \(v(x)\), and to that, add \(u(x)\) multiplied by the derivative of \(v(x)\). This rule is very handy and simplifies the process significantly.
The Chain Rule
The chain rule is another powerful tool in differentiation, particularly useful when dealing with composite functions. A composite function is a function made up of other functions, such as \(v(x) = \sin(2x)\). To differentiate a composite function \(h(g(x))\), use the chain rule:

\[\frac{d}{dx} h(g(x)) = h'(g(x)) \cdot g'(x)\]

For our function \(v(x) = \sin(2x)\), set \(h(u) = \sin(u)\) and \(g(x) = 2x\). Thus, applying the chain rule:

\[v'(x) = \frac{d}{dx} \sin(2x) = \cos(2x) \cdot 2 = 2 \cos(2x)\]
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. In simple terms, it鈥檚 the process of finding a derivative.

When differentiating functions, certain rules and techniques, like the product rule and the chain rule, come in handy to simplify the process and avoid errors.

Let's recap the steps to differentiate the function \(f(x) = 9x \sin(2x)\) using these rules:
  • Identify the functions \(u(x) = 9x\) and \(v(x) = \sin(2x)\).
  • Use the product rule to find \(f'(x)\).
  • Differentiate \(u(x)\) to find \(u'(x)\).
  • Apply the chain rule to differentiate \(v(x)\).
  • Combine the results using the product rule formula.
The final derivative is: \[f'(x) = 9 \sin(2x) + 18x \cos(2x)\]

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Most popular questions from this chapter

Consider the functions \(r(t)=t^{t}\) and \(s(t)=\arccos (t),\) for which you are given the facts that \(r^{\prime}(t)=t^{t}(\ln (t)+1)\) and \(s^{\prime}(t)=-\frac{1}{\sqrt{1-t^{2}}} .\) Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0

Let \(f(v)\) be the gas consumption (in liters/km) of a car going at velocity \(v\) (in km/hour). In other words, \(f(v)\) tells you how many liters of gas the car uses to go one kilometer if it is traveling at \(v\) kilometers per hour. In addition, suppose that \(f(80)=0.05\) and \(f^{\prime}(80)=\) \(0.0004 .\) a. Let \(g(v)\) be the distance the same car goes on one liter of gas at velocity \(v\). What is the relationship between \(f(v)\) and \(g(v)\) ? Hence find \(g(80)\) and \(g^{\prime}(80)\). b. Let \(h(v)\) be the gas consumption in liters per hour of a car going at velocity \(v\). In other words, \(h(v)\) tells you how many liters of gas the car uses in one hour if it is going at velocity \(v\). What is the algebraic relationship between \(h(v)\) and \(f(v)\) ? Hence find \(h(80)\) and \(h^{\prime}(80)\). c. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

Find the derivative of \(y=\sqrt{x}\left(x^{2}+4\right)\). \(\frac{d y}{d x}=\) __________

Let \(p(z)\) be given by the rule $$ p(z)=\frac{z \tan (z)}{z^{2} \sec (z)+1}+3 e^{z}+1. $$ a. Determine \(p^{\prime}(z)\) b. Find an equation for the tangent line to \(p\) at the point where \(z=0\). c. At \(z=0\), is \(p\) increasing, decreasing, or neither? Why?

Let \(f(x)=\sin (x) \cot (x)\). a. Use the product rule to find \(f^{\prime}(x)\). b. True or false: for all real numbers \(x, f(x)=\cos (x)\). c. Explain why the function that you found in (a) is almost the opposite of the sine function, but not quite. (Hint: convert all of the trigonometric functions in (a) to sines and cosines, and work to simplify. Think carefully about the domain of \(f\) and the domain of \(\left.f^{\prime} .\right)\).
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