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Chapter 4: Problem 33

Find the derivative with respect to the independent variable. $$ g(t)=\frac{\sin (2 t)+1}{\cos (6 t)-1} $$

Short Answer

Expert verified
The derivative is \( \frac{2\cos(2t)\cos(6t) - 2\cos(2t) + 6\sin(6t)\sin(2t) + 6\sin(6t)}{(\cos(6t) - 1)^2} \).

Step by step solution

01

Identify the Rule to Use

We need to find the derivative of a quotient of functions. The appropriate rule to use here is the quotient rule, which is given by: \ \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] \ where \( u = \sin(2t) + 1 \) and \( v = \cos(6t) - 1 \).
02

Differentiate the Numerator

The numerator function is \( u(t) = \sin(2t) + 1 \). To differentiate, use the chain rule. \ The derivative is: \ \[ \frac{du}{dt} = 2 \cos(2t) \] \ This is because the derivative of \( \sin(2t) \) is \( 2\cos(2t) \) and the constant 1 has a derivative of zero.
03

Differentiate the Denominator

The denominator function is \( v(t) = \cos(6t) - 1 \). Differentiating using the chain rule gives: \ \[ \frac{dv}{dt} = -6 \sin(6t) \] \ Here, the derivative of \( \cos(6t) \) is \( -6\sin(6t) \).
04

Apply the Quotient Rule

Using the quotient rule and substituting in the derivatives from Steps 2 and 3: \ \[ \frac{d}{dt} \left( \frac{\sin(2t) + 1}{\cos(6t) - 1} \right) = \frac{(\cos(6t) - 1)(2\cos(2t)) - (\sin(2t) + 1)(-6\sin(6t))}{(\cos(6t) - 1)^2} \] \ Simplifying gives us the derivative of the function \( g(t) \).
05

Simplify the Expression

The expression for the derivative is: \ \[ \frac{2\cos(2t) (\cos(6t) - 1) + 6 \sin(6t)(\sin(2t) + 1)}{(\cos(6t) - 1)^2} \] \ Multiply through: \ \[ = \frac{2\cos(2t)\cos(6t) - 2\cos(2t) + 6\sin(6t)\sin(2t) + 6\sin(6t)}{(\cos(6t) - 1)^2} \] \ This is the simplified form of the derivative of \( g(t) \).

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Key Concepts

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

Quotient Rule
When dealing with the derivative of a fraction, where you have one function divided by another, the quotient rule is the right tool for the job. This rule helps us find the derivative of two functions in a numerator and denominator relationship.
The formula for the quotient ruleis:
  • \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
Here, \(u\) is the function in the numerator, and \(v\) is the function in the denominator.
So, if you ever need to find a derivative of a function expressed as a fraction, remember these:
  • Multiply the derivative of the top by the bottom.
  • Subtract the top times the derivative of the bottom.
  • Then, divide everything by the bottom squared.
It may sound a bit complex, but it really boils down to following the formula and being careful with your calculations.
Chain Rule
In calculus, the chain rule is a method for finding the derivative of composite functions, or functions inside of functions.
Think of it like peeling an onion, where you must deal with each layer individually until you get to the core.
The chain rule formula is:
  • \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
Let's break this down. If you have two functions composed together, like \(f(g(t))\), you'd find the derivative of the outer function \(f\), leaving the inner function alone. Then, you multiply it by the derivative of the inner function \(g\).
With the original problem, this is why we handled \(\sin(2t)\) as \(2 \cos(2t)\), and \(\cos(6t)\) as \(-6 \sin(6t)\). You take the standard derivative of the trigonometric function and multiply by the derivative of the inner function's argument.
Trigonometric Functions
When working with derivatives of trigonometric functions such as \(\sin\) and \(\cos\), it's vital to remember their unique derivative rules.
These derivatives often pop up in calculus, so let's tackle the basics.
  • The derivative of \(\sin(x)\) is \(\cos(x)\).
  • The derivative of \(\cos(x)\) is \(-\sin(x)\).
Notice how the function changes from \(\sin\) to \(\cos\), or vice versa, and also pay attention to the negative sign associated with \(\cos\).
This is crucial when using these derivatives, especially in conjunction with the chain rule for functions like \(\sin(2t)\) or \(\cos(6t)\) that have more than a simple \(t\) inside them.
Keeping these rules in mind makes calculating derivatives much easier, even as the problems grow in complexity.

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Most popular questions from this chapter

A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x| .\) \(f(x)=1-3 x, x=-2 \pm 0.3\)

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=(1-x)^{-n}\) at \(a=0 .\) (Assume that \(n\) is a positive integer.)

Suppose that the concentration of nitrogen in a lake exhibits periodic behavior. That is, if we denote the concentration of nitrogen at time \(t\) by \(c(t)\), then we assume that $$c(t)=2+\sin \left(\frac{\pi}{2} t\right)$$ (a) Find \(\frac{d c}{d t}\). (b) Use a graphing calculator to graph both \(c(t)\) and \(\frac{d c}{d t}\) in the same coordinate system. (c) By inspecting the graph in (b), answer the following questions: (i) When \(c(t)\) reaches a maximum, what is the value of \(d c / d t ?\) (ii) When \(d c / d t\) is positive, is \(c(t)\) increasing or decreasing? (iii) What can you say about \(c(t)\) when \(d c / d t=0 ?\)

Find the derivative at the indicated point from the graph of \(y=f(x)\). \(f(x)=2 x-3 ; x=-1\)

The growth rate of a fungus varies over the course of one day. You find that the size of the fungus is given as a function of time by: $$L(t)=3.6 t+1.2 \cos (2 \pi t / 24)$$ where \(t\) is the time in hours, and \(L(t)\) is the size in millimeters. (a) Calculate the growth rate \(d L / d t\) (b) What is the largest growth rate of the fungus? What is the smallest growth rate?
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