Problem 48 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$q=\cot \left(\frac{\sin t}{t}\right)$$

Step-by-Step Solution

Answer

The derivative is \(-\csc^2\left(\frac{\sin t}{t}\right) \frac{(t\cos t - \sin t)}{t^2}\)."

Step 1: Identify the Outer Function

The given function is \( q = \cot \left(\frac{\sin t}{t}\right) \). The outer function is \( \cot x \), where \( x = \frac{\sin t}{t} \). We'll need the derivative of \( \cot x \), which is \( -\csc^2 x \).

Step 2: Differentiate the Inner Function

Next, we need to differentiate the inner function \( x = \frac{\sin t}{t} \) with respect to \( t \). Apply the quotient rule: \[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \] where \( u = \sin t \) and \( v = t \).

Step 3: Apply the Quotient Rule

Calculate the derivative using the quotient rule: \( u = \sin t \) (derivative \( \cos t \)) and \( v = t \) (derivative \( 1 \)). Thus, \[ \frac{d}{dt}\left(\frac{\sin t}{t}\right) = \frac{t \cdot \cos t - \sin t \cdot 1}{t^2} = \frac{t\cos t - \sin t}{t^2}. \]

Step 4: Combine Derivatives Using the Chain Rule

Use the chain rule to find \( \frac{dq}{dt} \). The derivative of the outer function is \( -\csc^2\left(\frac{\sin t}{t}\right) \), and the derivative of the inner function is \( \frac{t\cos t - \sin t}{t^2} \). Therefore, \[ \frac{dq}{dt} = -\csc^2\left(\frac{\sin t}{t}\right) \cdot \frac{t\cos t - \sin t}{t^2}. \]

Step 5: Simplify the Expression

Write the final derivative expression neatly: \[ \frac{dq}{dt} = -\csc^2\left(\frac{\sin t}{t}\right) \frac{(t\cos t - \sin t)}{t^2}. \]

Key Concepts

Quotient Rule

When you have a function formed by dividing one function by another, like in our exercise with the inner function \( \frac{\sin t}{t} \), you apply the quotient rule to find its derivative. The quotient rule is crucial in different calculus problems where you deal with ratios of two differentiable functions.

The formula for the quotient rule is: \ \[\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \ \]

• u is the numerator function. In our case, it’s \( \sin t \).
• v is the denominator function. Here, it’s \( t \).

To differentiate \( \frac{\sin t}{t} \), first calculate the derivatives of \( u \) and \( v \).

• \( \frac{du}{dt} = \cos t \) since the derivative of \( \sin t \) is \( \cos t \).
• \( \frac{dv}{dt} = 1 \) since the derivative of \( t \) is \( 1 \).

Substituting into the quotient rule gives us: \ \[\frac{t \cdot \cos t - \sin t \cdot 1}{t^2} = \frac{t \cos t - \sin t}{t^2} \ \]And that's how the quotient rule is applied to find the derivative of a division of two functions.

Chain Rule

The chain rule is all about taking derivatives of composite functions. In this problem, the function is composite because it consists of an outer function \( \cot x \) and an inner function \( x = \frac{\sin t}{t} \).

The chain rule helps you determine the derivative of such functions by processing them layer by layer. Simply put, differentiate the outer function and multiply it by the derivative of the inner function.

Use the formula: \ \[\frac{d}{dt}[\text{outer function (inner function)}] = \text{outer derivative} \cdot \text{inner derivative}\ \]

• Here, the outer function is \( \cot(x) \) with a derivative of \( -\csc^2(x) \).
• We previously calculated the derivative of the inner function as \( \frac{t \cos t - \sin t}{t^2} \).

Substituting the derivatives into the chain rule formula, you find the derivative: \ \[\frac{dq}{dt} = -\csc^2\left(\frac{\sin t}{t}\right) \cdot \frac{t\cos t - \sin t}{t^2} \ \]The chain rule effectively combines the derivative processes in an organized way allowing you to handle complex functions smoothly.

Trigonometric Functions Differentiation

Understanding how to differentiate trigonometric functions is essential in calculus, especially for more complex equations involving these functions. In our exercise, the function \( q = \cot \left( \frac{\sin t}{t} \right) \) involves the cotangent function, which is a trigonometric function.

Trigonometric derivatives are fundamental and include:

• The derivative of \( \sin t \) is \( \cos t \).
• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( \tan t \) is \( \sec^2 t \).
• The derivative of \( \cot t \) is \( -\csc^2 t \).
• The derivative of \( \sec t \) is \( \sec t \tan t \).
• The derivative of \( \csc t \) is \( -\csc t \cot t \).

In our problem, knowing \( \cot t = \frac{1}{\tan t} \), and its derivative \( -\csc^2 t \) is directly used to solve it.
Each trigonometric function follows set differentiation rules, which you apply depending on the given variable or angle in the function.

Having memorized these basic derivatives helps you compute these problems faster and be more efficient in approaching similar calculus challenges.