Problem 32 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cot ^{-1} \sqrt{t-1}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dt} = -\frac{1}{2t\sqrt{t-1}}\)

Step 1: Identify the function and the variable

The exercise requires us to find the derivative of \(y\) with respect to \(t\). The function given is \(y = \cot^{-1} \sqrt{t-1}\).

Step 2: Apply the chain rule

To find the derivative \(\frac{dy}{dt}\), we need to apply the chain rule. This rule states that if a function \(y = f(u)\) and \(u = g(t)\), then \(\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}\). Here \(u = \sqrt{t-1}\) and \(y = \cot^{-1}(u)\).

Step 3: Derivative of cotangent inverse

The derivative of \(y = \cot^{-1}(u)\) with respect to \(u\) is \(\frac{dy}{du} = -\frac{1}{1+u^2}\).

Step 4: Derivative of the square root function

The derivative of \(u = \sqrt{t-1}\) with respect to \(t\) is \(\frac{du}{dt} = \frac{1}{2\sqrt{t-1}}\).

Step 5: Multiply derivatives using the chain rule

Combining these results using the chain rule, we have: \[\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt} = -\frac{1}{1+u^2} \times \frac{1}{2\sqrt{t-1}}\]

Step 6: Substitute back for \(u\) and simplify

Substitute \(u = \sqrt{t-1}\) back into the equation: \[\frac{dy}{dt} = -\frac{1}{1+(\sqrt{t-1})^2} \times \frac{1}{2\sqrt{t-1}} = -\frac{1}{1+(t-1)} \times \frac{1}{2\sqrt{t-1}}\] This simplifies to: \[\frac{dy}{dt} = -\frac{1}{2t\sqrt{t-1}}\]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are fundamental in calculus and trigonometry. They essentially reverse the roles of the regular trigonometric functions, allowing us to find angles given a ratio. In this exercise, we focus on the inverse cotangent function, written as \( \cot^{-1}(x) \).

• The inverse cotangent function, \( \cot^{-1}(x) \), returns the angle whose cotangent is \( x \).
• One important property is its derivative: \( \frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1 + x^2} \).

This derivative indicates how the angle changes with respect to \( x \). The negative sign means that as \( x \) increases, the angle decreases.

Chain Rule

The chain rule is a powerful differentiation technique used when dealing with composite functions. It allows us to take the derivative of a function that is nested inside another function. For our problem, the function \( y = \cot^{-1}(\sqrt{t-1}) \) involves two layers:

• The outer layer is \( \cot^{-1}(u) \), where \( u = \sqrt{t-1} \).
• The inner layer is \( \sqrt{t-1} \), which depends on \( t \).

To find the overall derivative, we differentiate the outer layer with respect to the inner function, and then multiply by the derivative of the inner function itself. This step-by-step approach simplifies the complex differentiation process.

Differentiation Techniques

Differentiation techniques are critical tools in understanding rates of change. For our problem, we utilized several key techniques: inverse trigonometric differentiation and square root differentiation.

• The derivative of \( \cot^{-1}(u) \) is \( -\frac{1}{1+u^2} \).
• The derivative of a square root function \( \sqrt{t-1} \) is calculated using the formula \( \frac{1}{2\sqrt{t-1}} \).

By combining these techniques using the chain rule, we successfully differentiated the composite function. This combination highlights the importance of understanding different differentiation methods to tackle complex calculus problems.