Problem 67 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\tan ^{2}\left(\sin ^{3} t\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = 6 \tan(\sin^3 t) \sec^2(\sin^3 t) \sin^2(t) \cos(t) \)

Step 1: Identify the Outer Function

The given function is composite. We start by identifying the outermost function. Here, the outer function is \( y = u^2 \) where \( u = \tan(\sin^3 t) \). This suggests using the chain rule.

Step 2: Differentiate the Outer Function

Differentiate \( y = u^2 \) with respect to \( u \):\[ \frac{dy}{du} = 2u \]

Step 3: Identify the Intermediate Function

Set \( u = \tan(v) \) where \( v = \sin^3 t \). We will need to find \( \frac{du}{dv} \).

Step 4: Differentiate the Intermediate Function

Differentiate \( u = \tan(v) \) with respect to \( v \):\[ \frac{du}{dv} = \sec^2(v) \]

Step 5: Identify the Innermost Function

The innermost function is \( v = \sin^3 t \). Differentiate this function with respect to \( t \) using the power and chain rules.

Step 6: Differentiate the Innermost Function

Differentiate \( v = \sin^3 t \):\[ \frac{dv}{dt} = 3\sin^2(t) \cdot \cos(t) \]

Step 7: Apply the Chain Rule

Using the chain rule:\[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} \]Substitute the values from previous steps:\[ \frac{dy}{dt} = 2u \cdot \sec^2(v) \cdot 3\sin^2(t) \cdot \cos(t) \]

Step 8: Substitute Back Variables

Substitute back the values of \( u \) and \( v \):\[ u = \tan(\sin^3 t) \]\[ v = \sin^3 t \]Resulting in:\[ \frac{dy}{dt} = 2 \tan(\sin^3 t) \cdot \sec^2(\sin^3 t) \cdot 3\sin^2(t) \cdot \cos(t) \]

Step 9: Simplify the Expression

Simplify the expression if possible:\[ \frac{dy}{dt} = 6 \tan(\sin^3 t) \sec^2(\sin^3 t) \sin^2(t) \cos(t) \]

Key Concepts

Chain Rule

The chain rule is an essential part of differentiation, especially when dealing with composite functions. It provides a method to find the derivative of a function that contains another function within it. When using the chain rule, we take the derivative of the outer function and multiply it by the derivative of the inner function.
To understand how this works with our example, consider the composite function given:

• The outer function is the square of a tangent function: \( y = u^2 \).
• It helps us identify which functions to differentiate step by step.

By following the steps of the chain rule, we systematically differentiate each layer of the function until we get the derivative \( \frac{dy}{dt} \). Each component of the function is tackled individually, which makes differentiating complex functions manageable.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, frequently appear in calculus problems. These functions have their derivatives, essential for solving problems like the one at hand. In the given exercise, we encounter \( \tan(x) \), a trigonometric function with its own unique derivative.
The derivative of \( \tan(x) \) is \( \sec^2(x) \), which we use in our solution. Understanding the derivatives of trigonometric functions is crucial because they reveal how these functions change concerning time or angle.

• Knowing these derivatives helps us apply them correctly in solving calculus problems.
• Each derivative connects back to their fundamental trigonometric identities.

These links allow us to solve complex differentiation problems by applying known principles about these functions.

Composite Functions

Composite functions are functions within functions, making them a bit tricky to differentiate. They require more layers of differentiation, hence the importance of applying the chain rule. Let's break down our exercise function as an example:

• The innermost function, \( v = \sin^3(t) \), requires using the chain rule within itself.
• This function is plugged into the the tangent function: \( u = \tan(v) \).

Following through: When we differentiate each set of functions, we simplify the overall function by focusing on one section at a time. This turns an otherwise difficult problem into something manageable, using structured steps to reach a comprehensive derivative. Understanding how to work with composite functions builds a solid foundation for handling complex calculus problems.

Calculus Problem Solving

Solving calculus problems often involves piecing together multiple concepts, such as the chain rule and trigonometric identities, as seen in this exercise. To solve any calculus problem, it helps to:

• Identify and separate functions into manageable parts.
• Apply the right differentiation rules, like the chain rule or trigonometric derivatives.
• Substitute and simplify as needed to reach a concise solution.

Each step taken in calculus problem-solving is a piece of a larger puzzle. The key is understanding which rules apply at which stage, and how simplification helps bring out the core relationship between variables and their rates of change.
In our derivative problem, we tackled a complex function in stages. By using differentiation rules and simplifying results, we found \( \frac{dy}{dt} = 6 \tan(\sin^3 t) \sec^2(\sin^3 t) \sin^2(t) \cos(t) \). This illustrates the strategic nature of calculus problem-solving.