Problem 54 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=(1+\cot (t / 2))^{-2}$$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dt} = (1 + \cot(t/2))^{-3} \cdot \csc^2(t/2) \).

Step 1: Identify the function form

The given function is \( y = (1 + \cot(t/2))^{-2} \). This is a composition of functions where \( y = (1 + u)^{-2} \) with \( u = \cot(t/2) \). We need to apply the chain rule in combination with the derivative of a trigonometric function.

Step 2: Differentiate the outer function

First, differentiate \( y = (1 + u)^{-2} \) with respect to \( u \). This is of the form \( f(u) = (1+u)^{-2} \). The derivative \( f'(u) = -2(1+u)^{-3} \).

Step 3: Differentiate the inner function

Next, differentiate \( u = \cot(t/2) \) with respect to \( t \). The derivative of \( \cot(x) \) is \( -\csc^2(x) \), so \( \frac{d}{dt} \left( \cot(t/2) \right) = -\csc^2(t/2) \cdot \frac{1}{2} \).

Step 4: Apply the chain rule

According to the chain rule, \( \frac{dy}{dt} = f'(u) \cdot \frac{du}{dt} \). So, substituting in the expressions: \[ \frac{dy}{dt} = -2(1 + \cot(t/2))^{-3} \cdot (-\frac{1}{2} \csc^2(t/2)) \].

Step 5: Simplify the expression

Simplifying the expression from Step 4 gives: \[ \frac{dy}{dt} = (1 + \cot(t/2))^{-3} \cdot \csc^2(t/2) \].

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus used for differentiating composite functions. It helps us find the derivative of one function that is nested within another. In simpler terms, if you have a function within a function, like peeling an onion layer by layer, you use the chain rule to differentiate it.
The chain rule is expressed as: If you have a function \( y = f(g(x)) \), then its derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This formula shows that you first take the derivative of the outer function concerning the inner function, and then multiply it by the derivative of the inner function with respect to \( x \).
In our exercise, we identified the function \( y = (1 + \cot(t/2))^{-2} \) as a composition where the outer function is \((1 + u)^{-2}\) and the inner function is \(u = \cot(t/2)\). The chain rule allows us to differentiate each part step by step.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are the building blocks of trigonometry. These functions relate the angles of triangles to the lengths of their sides and are pivotal in calculus, especially when involving derivatives.
In our specific problem, we dealt with the cotangent function \( \cot(x) \), which can be defined as the reciprocal of the tangent function:

• \( \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \).

When differentiating \( \cot(x) \), we use the fact that:

• \( \frac{d}{dx}(\cot(x)) = -\csc^2(x) \).

The derivative of \( \cot \) gave us the negative cosecant squared, essential in calculating the rate of change of our inner function \( u = \cot(t/2) \) with respect to \( t \).

Calculus Exercises

Calculus exercises often involve applying different rules and formulas to solve problems. This can include differentiation, integration, and understanding the behavior of functions.
When tackling calculus exercises, it's important to:

• Identify the form of functions involved, such as polynomial, trigonometric, or exponential.
• Decide the appropriate calculus rules, like the chain rule, product rule, or quotient rule.
• Calculate derivatives accurately and simplify the expressions for comprehensive solutions.

In this particular exercise, we combined trigonometric identities and the chain rule steps to differentiate the function. Breaking it down into manageable steps enabled us to find the derivative \(\frac{dy}{dt}\) correctly. Simplifying the final expression ensures clarity and precision in presenting the solution.