Problem 30 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1} \frac{3}{t^{2}}$$

Step-by-Step Solution

Answer

\(-\frac{6}{t \sqrt{t^4 - 9}}\)

Step 1: Identify the function and variable

The function given is \(y = \sin^{-1}\left(\frac{3}{t^2}\right)\). The goal is to find \(\frac{dy}{dt}\), the derivative of \(y\) with respect to \(t\).

Step 2: Differentiate the inverse sine function

Use the derivative formula for the inverse sine function: if \(y = \sin^{-1}(u)\), then \(\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}}\). Here, \(u = \frac{3}{t^2}\).

Step 3: Differentiate u with respect to t

Let \(u = \frac{3}{t^2}\). Differentiate \(u\) with respect to \(t\):\[ u = \frac{3}{t^2} = 3t^{-2} \]\[ \frac{du}{dt} = \frac{d}{dt}(3t^{-2}) = -6t^{-3} \]

Step 4: Apply the chain rule

Use the chain rule to find \(\frac{dy}{dt}\):\[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = \frac{1}{\sqrt{1-\left(\frac{3}{t^2}\right)^2}} \cdot (-6t^{-3}) \]

Step 5: Simplify the expression

First, simplify \(1-\left(\frac{3}{t^2}\right)^2\):\(\left(\frac{3}{t^2}\right)^2 = \frac{9}{t^4}\)\[ 1 - \frac{9}{t^4} = \frac{t^4 - 9}{t^4} \]Now, substitute into the derivative:\[ \frac{dy}{dt} = -6t^{-3} \cdot \frac{1}{\sqrt{\frac{t^4 - 9}{t^4}}} = -6t^{-3} \cdot \frac{t^2}{\sqrt{t^4 - 9}} = -\frac{6t^{-1}}{\sqrt{t^4 - 9}} \]

Step 6: Final Expression

The derivative of \(y = \sin^{-1}\left(\frac{3}{t^2}\right)\) with respect to \(t\) is\[ \frac{dy}{dt} = -\frac{6}{t \sqrt{t^4 - 9}} \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the action of standard trigonometric functions. In this exercise, we use the inverse sine function, denoted as \( \sin^{-1}(x) \). This function helps us find an angle whose sine is a particular value.
For example, if \( y = \sin^{-1}(u) \), then this equation means that the sine of angle \( y \) is \( u \). The derivative of the inverse sine function is important when dealing with differentiation problems involving trigonometric relationships. It is given by:

• \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \)

In this problem, we applied this concept to find the derivative of \( y = \sin^{-1}\left(\frac{3}{t^2}\right) \), which involves a bit more complexity due to the term \( \frac{3}{t^2} \). As described in the steps, handling such functions entails not only differentiating the inverse function but also maintaining the correctness of the inner function, \( u \). Understanding this derivative is key for the successful differentiation of composite functions involving inverse trigonometry.

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of a composite function. When dealing with functions nested within each other, like \( y = \sin^{-1}\left(\frac{3}{t^2}\right) \), the chain rule provides a structured way to differentiate such functions. Simply put, it tells us to "chain" the derivatives together.
The rule is often stated as:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

Here, we apply the chain rule to differentiate the composite function \( y = \sin^{-1}\left(\frac{3}{t^2}\right) \). We first differentiated the outer function (the inverse sine) with respect to its inner function \( u = \frac{3}{t^2} \), and then the inner function \( u \) with respect to \( t \). This sequence allowed us to compute:

• \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \)

The chain rule is essential for solving complex calculus problems efficiently and forms the backbone of many differentiation tasks.

Differentiation

Differentiation is a core concept in calculus dealing with how functions change. It allows us to find the rate at which one variable changes concerning another. When differentiating, we calculate derivatives, which are expressions that sketch the instantaneous rate of change or the slope of the curve at any point on the graph.
In the exercise, the process of differentiation was applied to the inverse trigonometric function \( y = \sin^{-1}\left(\frac{3}{t^2}\right) \). This required several steps and techniques such as the inverse function derivative and the chain rule. By differentiating such expressions, you can determine how involved variables interact and influence each other.
Steps to differentiate effectively include:

• Recognizing the function and identifying necessary rules like inverse function derivatives.
• Applying the chain rule to handle composite and multi-layered functions.
• Performing algebraic manipulation to simplify the expressions to a manageable form.

Practicing differentiation with a variety of functions makes this skill invaluable in solving real-world problems modelled in mathematical terms.