Problem 44 - Thomas Calculus

Question

In Exercises \(39-48,\) find \(d y / d t\) $$ y=\cos \left(5 \sin \left(\frac{t}{3}\right)\right) $$

Step-by-Step Solution

Answer

\(-\frac{5}{3} \sin\left(5 \sin\left(\frac{t}{3}\right)\right) \cdot \cos\left(\frac{t}{3}\right)\) is the derivative.

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \cos \left(5 \sin \left(\frac{t}{3}\right)\right) \) with respect to \( t \). This requires using the chain rule multiple times due to the nested functions.

Step 2: Apply the Outer Chain Rule

Identify the outermost function, which is \( \cos(u) \). The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \). Set \( u = 5 \sin \left(\frac{t}{3}\right) \). Then the derivative is \( -\sin(u) \cdot \frac{du}{dt} \).

Step 3: Differentiate the Intermediate Function

Now, differentiate \( u = 5 \sin \left(\frac{t}{3}\right) \) with respect to \( t \). This requires the chain rule again. Identify the outer function as \( 5\sin(v) \) where \( v = \frac{t}{3} \). The derivative of \( 5\sin(v) \) with respect to \( v \) is \( 5\cos(v) \).

Step 4: Differentiate the Innermost Function

Differentiate \( v = \frac{t}{3} \) with respect to \( t \), which is simply \( \frac{1}{3} \).

Step 5: Combine the Derivatives

Combine the derivatives from each step using the chain rule:\[ \frac{dy}{dt} = -\sin\left(5 \sin\left(\frac{t}{3}\right)\right) \cdot 5\cos\left(\frac{t}{3}\right) \cdot \frac{1}{3} \]This simplifies to:\[ \frac{dy}{dt} = -\frac{5}{3} \sin\left(5 \sin\left(\frac{t}{3}\right)\right) \cdot \cos\left(\frac{t}{3}\right) \]

Key Concepts

Understanding Derivatives

When dealing with calculus, derivatives are crucial. They measure how a function changes as its input changes. In simpler terms, a derivative tells you the rate at which one quantity changes with respect to another. For example, if you have a position function, its derivative with respect to time gives you velocity.
In the context of our exercise, the goal is to find the derivative of a complex function with respect to a variable, usually time or space. The function in the problem, \( y = \cos \left(5 \sin \left(\frac{t}{3}\right)\right) \), is complex due to multiple layers of functions. Each inner function influences the outer one, and we use derivatives to explore these changes. These types of problems often involve the chain rule, which helps us differentiate composite functions effectively. This brings us to our next topic: the trigonometric functions involved.

Trigonometric Functions in Derivatives

Trigonometric functions like sine, cosine, and tangent frequently appear in calculus. They have unique properties that apply to their derivatives.
For instance, when differentiating the cosine function \( \cos(u) \), the result is \( -\sin(u) \). Similarly, for \( \sin(u) \), the derivative is \( \cos(u) \). These rules are essential when applying the chain rule, especially in cases with multiple trigonometric layers.In our exercise, the function \( y = \cos \left(5 \sin \left(\frac{t}{3}\right)\right) \) contains both cosine and sine:

• The outermost function uses \( \cos \), requiring its derivative, \( -\sin(u) \).
• The intermediate function has \( 5 \sin(v) \), where \( v = \frac{t}{3} \). Here, the derivative is \( 5 \cos(v) \).

By understanding these trigonometric derivatives, we can strategically apply the chain rule to our problem. Next, we address nested functions.

Nested Functions and the Chain Rule

Nested functions are simply functions within functions. When dealing with nested functions, determining derivatives can seem daunting. However, the chain rule simplifies this process.
The chain rule allows us to differentiate composite functions by addressing each function individually, one layer at a time.In the given problem, the function's layers are:

• Outermost: \( \cos(u) \) where \( u = 5 \sin \left(\frac{t}{3}\right) \)
• Intermediate: \( 5 \sin(v) \) where \( v = \frac{t}{3} \)
• Innermost: \( \frac{t}{3} \)

To find \( \frac{dy}{dt} \), we differentiate each layer:

• First, \( \cos(u) \) becomes \( -\sin(u) \cdot \frac{du}{dt} \).
• Next, \( 5 \sin(v) \) becomes \( 5 \cos(v) \cdot \frac{dv}{dt} \).
• Finally, \( \frac{t}{3} \) results in the derivative \( \frac{1}{3} \).

By combining these derivatives, we effectively use the chain rule to break down and solve for \( \frac{dy}{dt} \), which showcases the powerful utility of handling nested and composite functions.