Problem 33 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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 1: 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 \( v = \cos(6t) - 1 \).

Step 2: 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.

Step 3: 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 \( -6\sin(6t) \).

Step 4: 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 \( g(t) \).

Step 5: 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 \( g(t) \).

Key Concepts

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.