Problem 24 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$y=\cos ^{-1}\left(\sin ^{-1} t\right)$$

Step-by-Step Solution

Answer

The derivative is \(-\frac{1}{1 - t^2}\).

Step 1: Recognize the composition of functions

The function given is a composite function involving inverse trigonometric functions: \[ y = \cos^{-1}(\sin^{-1} t) \].This means we need to apply the chain rule to find its derivative, as it involves an outer function \(\cos^{-1}(x)\) and an inner function \(\sin^{-1}(t)\).

Step 2: Differentiate the outer function

The outer function is \(u = \cos^{-1}(x)\). The derivative of \(u\) with respect to \(x\) is given by: \[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}} \].In our context, \(x\) is actually \(\sin^{-1}(t)\).

Step 3: Differentiate the inner function

The inner function is \(x = \sin^{-1}(t)\). The derivative of \(x\) with respect to \(t\) is given by: \[ \frac{dx}{dt} = \frac{1}{\sqrt{1 - t^2}} \].

Step 4: Apply the chain rule

Using the chain rule, which states \(\frac{dy}{dt} = \frac{du}{dx} \cdot \frac{dx}{dt}\), we can now find the derivative of the composite function:\[\frac{dy}{dt} = \left( -\frac{1}{\sqrt{1 - (\sin^{-1} t)^2}} \right) \cdot \frac{1}{\sqrt{1 - t^2}}.\]

Step 5: Simplify the expression

We know from trigonometric identities that:\[1 - (\sin^{-1} t)^2 = \cos^2(\theta),\] where \(\theta = \sin^{-1}(t)\) and \(\cos(\theta) = \sqrt{1 - t^2}\).Therefore, the expression becomes:\[-\frac{1}{\sqrt{\cos^2(\theta)}} \cdot \frac{1}{\sqrt{1 - t^2}} = -\frac{1}{\sqrt{1 - t^2}} \cdot \frac{1}{\sqrt{1 - t^2}} = -\frac{1}{1 - t^2}.\]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles when given certain trigonometric values. They reverse the action of the standard trigonometric functions. For example, \(\sin^{-1}(t)\) gives the angle whose sine is \(t\). Inverse trigonometric functions include \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\).
These functions are crucial in problems where the original trigonometric function provides an obtuse or acute angle and we need to work back to find an angle given the function's output.
The domain and range of these functions are restricted because only certain values lead to real numerical angles. For instance, the domain of \(\sin^{-1}(x)\) is \(-1 \leq x \leq 1\) and its range is \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\). This limited range ensures that each input maps to a single output, making the function "inverse" to the original.

Chain Rule

The chain rule is a fundamental tool in calculus, used to differentiate composite functions. Composite functions are those that combine two or more functions, such as \(y = \cos^{-1}(\sin^{-1} t)\). The chain rule helps to find how changes in the inner variable affect the outer variable.
To apply the chain rule, understand the inner and outer functions. Given \( y = f(g(t)) \), the derivative is \((f(g(t)))' = f'(g(t)) \cdot g'(t)\).
In this exercise, the chain rule is applied by treating \(\cos^{-1}\) as the outer function and \(\sin^{-1}\) as the inner function. We find the derivative of \(\cos^{-1}(x)\) with respect to \(x\), and then the derivative of \(\sin^{-1}(t)\) with respect to \(t\).

• Outer Function Derivative: The derivative of \(\cos^{-1}(x)\) is \-\frac{1}{\sqrt{1-x^2}}\.
• Inner Function Derivative: The derivative of \(\sin^{-1}(t)\) is \ \frac{1}{\sqrt{1-t^2}}\.

Finally, multiply these derivatives to find the derivative of the entire function.

Trigonometric Identities

Trigonometric identities offer relationships between trigonometric functions and are essential in simplifying expressions. In this exercise, the Pythagorean identity is key. It says \(\sin^2(x) + \cos^2(x) = 1\).
This identity can be rearranged to find either the sine or cosine of an angle when the other is known:

• If \(\sin(x) = t\), then \(\cos(x) = \sqrt{1-t^2}\).

In the context of this problem, we use derivative simplification:

• Recognize that \(1 - (\sin^{-1}(t))^2 = \cos^2(\theta)\), where \(\theta\) is the angle given by \(\sin^{-1}(t)\).
• This leads to \(\cos(\theta) = \sqrt{1-t^2}\), aiding in expression simplification.

Ultimately, trigonometric identities make complex functions easier to differentiate, manipulate, and simplify by providing a direct path from trigonometric expressions to algebraic simplifications.