Problem 32 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ g(t)=\frac{\sin (3 t)}{\cos (5 t)} $$

Step-by-Step Solution

Answer

The derivative is \( g'(t) = \frac{\cos(2t)}{(\cos(5t))^2} \).

Step 1: Identify the function type

The function, \( g(t) = \frac{\sin (3t)}{\cos (5t)} \), is a quotient of two separate functions, which indicates it is suitable for applying the Quotient Rule.

Step 2: Recall the Quotient Rule

The Quotient Rule states that for a function \( h(x) = \frac{u(x)}{v(x)} \), the derivative \( h'(x) \) is given by: \[ h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] where \( u(x) = \sin(3t) \) and \( v(x) = \cos(5t) \).

Step 3: Differentiate the numerator

Differentiate \( u(t) = \sin(3t) \) with respect to \( t \). Using the chain rule, the derivative is \( u'(t) = 3\cos(3t) \).

Step 4: Differentiate the denominator

Differentiate \( v(t) = \cos(5t) \) with respect to \( t \). Using the chain rule, the derivative is \( v'(t) = -5\sin(5t) \).

Step 5: Apply the Quotient Rule

Substitute the derivatives found into the Quotient Rule formula: \[ g'(t) = \frac{(3\cos(3t))(\cos(5t)) - (\sin(3t))(-5\sin(5t))}{(\cos(5t))^2} \] which simplifies to: \[ g'(t) = \frac{3\cos(3t)\cos(5t) + 5\sin(3t)\sin(5t)}{(\cos(5t))^2} \].

Step 6: Simplify the expression

The derivative expression can be further simplified using the angle-sum identity \( \cos A \cos B + \sin A \sin B = \cos(A - B) \). Substitute into the formula: \[ g'(t) = \frac{\cos(3t - 5t)}{(\cos(5t))^2} = \frac{\cos(-2t)}{(\cos(5t))^2} \] which simplifies to \[ g'(t) = \frac{\cos(2t)}{(\cos(5t))^2} \] since \( \cos(-x) = \cos(x) \).

Key Concepts

Quotient Rule

The Quotient Rule is a technique used in calculus to find the derivative of a quotient of two functions. When faced with a function like \( g(t) = \frac{\sin(3t)}{\cos(5t)} \), the Quotient Rule provides a systematic way to differentiate.
Here’s how it works: if you have a function \( h(x) = \frac{u(x)}{v(x)} \), the derivative is:

• \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

In the example, \( u(t) = \sin(3t) \) is the numerator, and \( v(t) = \cos(5t) \) is the denominator. By identifying these, you can apply the Quotient Rule effectively. It's important to correctly differentiate both functions involved before applying the rule.
For instance, while applying the Quotient Rule to \( g(t) \), you calculate derivatives of both \( u(t) \) and \( v(t) \), then plug the results into the formula, remembering to always include the square of the denominator, \( [v(x)]^2 \), in the final expression.

Chain Rule

The Chain Rule is essential whenever you differentiate a composite function, like \( \sin(3t) \) or \( \cos(5t) \). These functions aren’t straightforward as they involve another function within. The Chain Rule tells us how to differentiate them properly. When you encounter a function \( f(g(x)) \), its derivative is found by:

• \( f'(g(x)) \cdot g'(x) \)

To apply this, differentiate the outer function while leaving the inside unchanged, then multiply by the derivative of the inside function.
In our example, differentiating \( \sin(3t) \) requires using the Chain Rule. You first identify \( f(t) = \sin(t) \) and \( g(t) = 3t \), then find their derivatives:

• Outer function derivative: \( \cos(3t) \)
• Inner function derivative: \( 3 \)

Multiply them to get \( 3\cos(3t) \). A similar step applies to \( \cos(5t) \), helping you derive the inside of another layer. The Chain Rule simplifies dealing with complex functions by systematically breaking them down.

Trigonometric Identities

Simplifying derivatives often involves using trigonometric identities. These identities can help you recognize patterns and simplify expressions. In our example, we saw use of the identity:

• \( \cos(A)\cos(B) + \sin(A)\sin(B) = \cos(A - B) \)

This was essential to simplify the expression obtained from differentiating \( g(t) = \frac{\sin(3t)}{\cos(5t)} \). Once the initial derivative was found through the Quotient Rule and it was: \[ g'(t) = \frac{3\cos(3t)\cos(5t) + 5\sin(3t)\sin(5t)}{(\cos(5t))^2} \] Applying the identity reduces \( 3t - 5t \), resulting in \( \cos(-2t) \), which simplifies to \( \cos(2t) \) since \( \cos(-x) = \cos(x) \).
Using such identities allows for easier manipulation and simplification, making trigonometric derivatives more manageable. When solving derivative problems, especially with trigonometric functions, being familiar with these identities simplifies your work and leads to more elegant solutions.