Problem 61 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\sin (\cos (2 t-5))$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = -2 \cos(\cos(2t-5)) \cdot \sin(2t-5) \)

Step 1: Understand Chain Rule

The function given is a composition of functions, which means we will apply the chain rule. The chain rule states that if you have a function within another function, such as \( y = f(g(t)) \), then the derivative is \( \frac{dy}{dt} = f'(g(t)) \cdot g'(t) \).

Step 2: Identify the Inner and Outer Functions

The outer function is \( f(u) = \sin(u) \), and the inner function is \( g(t) = \cos(2t - 5) \). We will first find derivatives for these functions.

Step 3: Differentiate the Outer Function

Differentiate the outer function \( f(u) = \sin(u) \). The derivative is \( f'(u) = \cos(u) \).

Step 4: Differentiate the Inner Function (Part 1)

The inner function is itself a composition. Start by differentiating \( h(v) = \cos(v) \), where \( v = 2t - 5 \). The derivative of \( \cos(v) \) is \( -\sin(v) \).

Step 5: Differentiate the Inner Function (Part 2)

Now take the derivative of \( v(t) = 2t - 5 \), which is simply \( v'(t) = 2 \).

Step 6: Apply the Chain Rule to Inner Function

Combine the results from Steps 4 and 5 to find the derivative of the entire inner function. \( g'(t) = h'(v) \cdot v'(t) = -\sin(2t - 5) \cdot 2 \).

Step 7: Apply the Chain Rule to the Entire Function

Now apply the chain rule to find \( \frac{dy}{dt} \) using \( f'(g(t)) \cdot g'(t) \). This yields \( \frac{dy}{dt} = \cos(\cos(2t - 5)) \cdot (-\sin(2t - 5) \cdot 2) \).

Step 8: Simplify the Derivative

Simplify the expression from Step 7, resulting in: \( \frac{dy}{dt} = -2 \cos(\cos(2t - 5)) \cdot \sin(2t - 5) \).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, used to differentiate composite functions. It helps find the derivative of a function that is nested within another function. Understanding this rule is crucial when dealing with functions like our given example, where we have layering of sine and cosine functions.

To apply the chain rule, first identify your functions: there will be an outer and an inner function. For instance, in the exercise where we have the function \(y = \sin(\cos(2t-5))\), the outer function is \(\sin(u)\), and the inner function is \(\cos(2t-5)\). Once these are identified, the chain rule can be expressed mathematically as follows:

• The derivative of \(f(g(t))\) is \(f'(g(t)) \times g'(t)\).

This combination allows us to differentiate even complex or layered functions systematically.

Composition of Functions

Composition of functions involves combining two or more functions such that the output of one function becomes the input of another. In calculus, this often involves functions inside other functions, like nested Russian dolls.

For the function \(y = \sin(\cos(2t-5))\), composition is clearly seen. The function \(\cos(2t-5)\) is nestled within \(\sin(u)\). This means you evaluate \(\cos(2t-5)\) first to get an angle, which then becomes the input for the sine function. Breaking it down:

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

Recognizing the composition helps set the stage for applying the chain rule correctly, as it delineates where one derivative starts and the next applies.

Derivative Calculation

Calculating derivatives involves finding how a function changes at any point, often representing rates of change. In the problem, once the chain rule is grasped, the actual derivative calculation follows.

Start with the derivatives of each function involved:

• First, take the derivative of the outer function: for \(f(u) = \sin(u)\), we have \(f'(u) = \cos(u)\).
• Next, focus on the inner function. Here, it is more layered: for \(g(t) = \cos(2t-5)\), first find the derivative of \(\cos(v)\), which is \(-\sin(v)\), and then multiply by the derivative of \(v(t) = 2t - 5\), which is \(2\).

Integrating these steps through the chain rule results in \( \frac{dy}{dt} = -2\cos(\cos(2t-5)) \cdot \sin(2t-5) \). This final step ensures that each small piece of the function's behavior is captured and combined correctly for a precise rate of change.