Problem 43 - Thomas Calculus

Question

In Exercises \(39-48,\) find \(d y / d t\) $$ y=\sin (\cos (2 t-5)) $$

Step-by-Step Solution

Answer

The derivative is \(-2 \cos(\cos(2t - 5)) \sin(2t - 5)\).

Step 1: Identify the Function

Given function to differentiate is \( y = \sin(\cos(2t - 5)) \). We need to find \( \frac{dy}{dt} \). This is a composite function, with sine and cosine functions nested inside one another.

Step 2: Apply the Chain Rule

To differentiate \( y = \sin(\cos(2t - 5)) \), use the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Step 3: Differentiate the Outer Function

The outer function is \( f(u) = \sin(u) \) where \( u = \cos(2t - 5) \). Therefore, the derivative of the outer function is \( f'(u) = \cos(u) \).

Step 4: Differentiate the Inner Function

We now need to differentiate the inner function \( u = \cos(2t - 5) \). The derivative of \( \cos(x) \) is \( -\sin(x) \). Therefore, the derivative with respect to \( t \) is \( u' = -\sin(2t - 5) \).

Step 5: Apply the Chain Rule to the Inner Argument

Inside \( \cos(2t - 5) \), there is another function, \( 2t - 5 \). The derivative of \( 2t - 5 \) with respect to \( t \) is \( 2 \).

Step 6: Combine Derivatives Using the Chain Rule

Combine derivatives using the chain rule: \( \frac{dy}{dt} = \cos(\cos(2t - 5)) \cdot (-\sin(2t - 5)) \cdot 2 \). Simplifying this gives, \[ \frac{dy}{dt} = -2 \cos(\cos(2t - 5)) \sin(2t - 5) \].

Step 7: Confirm the Final Derivative Expression

The final derivative expression is \( \frac{dy}{dt} = -2 \cos(\cos(2t - 5)) \sin(2t - 5) \).

Key Concepts

Composite Functions

Composite functions occur when one function is nested within another. In the problem of calculating the derivative of \( y = \sin(\cos(2t - 5)) \), there are actually multiple layers. The outer layer is the sine function, and inside it lies a cosine function. Finally, the expression \( 2t - 5 \) is buried within the cosine. As a student encountering composite functions, it is crucial to dissect these layers.

• Outer function: \( f(x) = \sin(x) \)
• Inner function: \( g(x) = \cos(2t - 5) \)
• Innermost function: \( h(t) = 2t - 5 \)

Understanding composite functions helps you use the chain rule more effectively, which is the primary tool needed here. Extracting each nested layer allows you to tackle its derivative one step at a time.

Trigonometric Differentiation

Differentiating trigonometric functions requires familiarity with their specific derivatives. Let’s look into how trigonometric differentiation works for this exercise, emphasizing the sine and cosine functions.

• The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \).
• For \( \cos(u) \), the derivative becomes \( -\sin(u) \).

In our example, since \( y = \sin(\cos(2t - 5)) \), the sine function wraps around the cosine function. For the outer sine function, its differentiation provides an inner cosine function, while the derivative of the inner cosine function presents a negative sine. Each function has its influence on the eventual derivative of the nested expression. Mastering these trigonometric rules makes it easier to handle more complex calculus problems involving such functions.

Calculus Problem Solving

Solving calculus problems often involves strategic application of rules and principles. In this exercise, particularly dealing with trigonometric composite functions, the chain rule becomes pivotal.To effectively solve such problems:1. **Break Down the Function**: Identify all constituent functions. 2. **Apply the Chain Rule**: Understand that for a composite function \( y = f(g(x)) \), the derivative is found by multiplying the derivative of the outer function by the derivative of the inner function: \[ \frac{dy}{dt} = f'(g(t)) \cdot g'(t). \]3. **Layer by Layer Differentiation**: For each inner function, continue the chain rule as each gets deeper nested, like peeling an onion.4. **Combine and Simplify Terms**: Once you have all the differentials, multiply them together and simplify, as shown by the step \( -2 \cos(\cos(2t - 5)) \sin(2t - 5) \).Through practice and careful application of these steps, you'll unravel the complexities of calculus problems and efficiently reach the solution.