Problem 62 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\cos \left(5 \sin \left(\frac{t}{3}\right)\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = -\frac{5}{3} \cdot \sin(5 \sin(\frac{t}{3})) \cdot \cos(\frac{t}{3}) \)."}

Step 1: Understand the Function Composition

The given function is a composition of multiple functions. First, there's a sine function applied to \(\frac{t}{3}\), then that result is passed into another sine function, and finally, the cosine function is applied. Thus, we have three layers of functions: \(y = \cos(u)\), where \(u = 5 \cdot \sin(v)\), and \(v = \frac{t}{3}\).

Step 2: Differentiate the Outer Function

Differentiate the outermost function using the chain rule. The derivative of \(y = \cos(u)\) with respect to \(u\) is \(-\sin(u)\). So, \( \frac{dy}{du} = -\sin(u)\).

Step 3: Differentiate the Middle Function

Now differentiate the middle function \(u = 5 \cdot \sin(v)\) with respect to \(v\). The derivative is \( \frac{du}{dv} = 5 \cdot \cos(v)\).

Step 4: Differentiate the Innermost Function

Differentiate the innermost function \(v = \frac{t}{3}\) with respect to \(t\). The derivative is \( \frac{dv}{dt} = \frac{1}{3}\).

Step 5: Apply the Chain Rule

To find \( \frac{dy}{dt}\), we apply the chain rule to combine all these derivatives: \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} = -\sin(u) \cdot 5 \cdot \cos(v) \cdot \frac{1}{3} \).

Step 6: Substitute the Expressions Back

Substitute \(u = 5 \sin(v)\) and \(v = \frac{t}{3}\) back into the expression: \( \frac{dy}{dt} = -\sin(5 \sin(\frac{t}{3})) \cdot 5 \cdot \cos(\frac{t}{3}) \cdot \frac{1}{3}\).

Step 7: Simplify the Expression

Simplify \( \frac{dy}{dt}\): \( \frac{dy}{dt} = -\frac{5}{3} \cdot \sin(5 \sin(\frac{t}{3})) \cdot \cos(\frac{t}{3}) \).

Key Concepts

Composition of Functions

When dealing with the composition of functions, you're essentially working with a function inside another function. This can be thought of like layers of an onion. In the given exercise, we have three layers: the outer cosine function, an inner sine function, and the innermost being a simple linear expression \( t/3 \).

• The innermost function, \( v = \frac{t}{3} \), is evaluated first, transforming \( t \).
• The result, \( v \), is then plugged into the trigonometric function \( u = 5 \cdot \sin(v) \), creating another layer.
• Finally, this result, \( u \), enters the final layer with the function \( y = \cos(u) \).

Understanding how these functions layer upon each other is crucial, especially when you apply differentiation rules like the chain rule.

Trigonometric Differentiation

Trigonometric differentiation deals with finding the derivatives of trigonometric functions, which are common in calculus problems. In this exercise, both the sine and cosine functions are involved.
**How to Differentiate Sine and Cosine:**

• The derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \).

For the given function \( y = \cos(5 \sin(\frac{t}{3})) \):

• We first differentiate the cosine function, resulting in \( -\sin(u) \).
• Then, the sine function inside needs differentiation, giving \( 5 \cdot \cos(v) \) where \( v = \frac{t}{3} \).

By carefully applying these rules, you can handle the trigonometric aspects of differentiation effectively.

Calculus Problem Solving

Solving calculus problems often involves recognizing patterns and functions within functions to properly apply differentiation rules. Let's break down this common calculus approach:
**Step-by-Step Application:**

• Identify all the functions within the problem statement.
• Differentiate from the outermost to the innermost function systematically using the chain rule.

In this example:

• We start with the differentiation of the outer cosine layer, using the chain rule to manage each subsequent layer.
• The process involves multiplying derivatives step by step: from the outermost, \( -\sin(u) \), to middle \( 5 \cdot \cos(v) \), and the inner \( \frac{1}{3} \).
• Finally, substitute back to ensure your expression represents the rate of change correctly.

By taking these organized steps, calculus problems become much more approachable and manageable, allowing for more accurate problem solving.