Problem 26 - Calculus Single Variable

Question

Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(-\pi / 2, \pi / 2) \rightarrow(-\infty, \infty), f(s)=\tan (s) $$

Step-by-Step Solution

Answer

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

Step 1: Recall the Inverse Function Derivative Rule

The Inverse Function Derivative Rule states that if \(y = f(x)\) is a differentiable function and \(f'(x) eq 0\), then the derivative of its inverse is \(\left(f^{-1}\right)'(y) = \frac{1}{f'(x)}\), where \(x = f^{-1}(y)\).

Step 2: Differentiate the Given Function

Given the function \(f(s) = \tan(s)\), we differentiate it with respect to \(s\). The derivative is \(f'(s) = \sec^2(s)\).

Step 3: Identify the Inverse Function

Since \(f(s) = \tan(s)\) and \(y = \tan(s)\), the inverse function is \(s = \tan^{-1}(y)\). Thus, \(s = f^{-1}(t)\).

Step 4: Apply the Inverse Function Derivative Rule

Using the inverse function derivative rule, we need to find \(\left(f^{-1}\right)'(t)\). For \(f(s) = \tan(s)\), the expression becomes \(\left(f^{-1}\right)'(t) = \frac{1}{\sec^2(s)}\), where \(s = \tan^{-1}(t)\).

Step 5: Simplify the Expression

Since \(\sec(s) = \frac{1}{\cos(s)}\), \(\sec^2(s) = \frac{1}{\cos^2(s)}\). Therefore, \(\left(f^{-1}\right)'(t) = \cos^2(s)\). By recognizing that \(\cos(s) = \frac{1}{\sqrt{1 + y^2}}\) from trigonometric identities, substituting \(y = t\) gives \(\cos^2(s) = \frac{1}{1+t^2}\).

Step 6: Final Expression for Inverse Derivative

Thus, the derivative of the inverse function \((f^{-1})^{\prime}(t)\) is \(\frac{1}{1+t^2}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus and trigonometry, providing relationships between various trigonometric functions. These identities can simplify complex expressions and are fundamental when working with derivatives, such as in finding the derivative of inverse functions.
One critical identity used in differentiation is:

• \( \sec^2(s) = 1 + \tan^2(s) \)

This identity helps in understanding the derivatives of functions involving the tangent and secant functions. By expressing secant terms in terms of tangent, we can simplify expressions and make calculus operations, like differentiation, more straightforward.
Using this identity, we can deduce relationships like the one needed for the inverse function derivative rule. Recognizing these trigonometric identities aids in manipulating and applying them to solve calculus problems efficiently.

Differentiation

Differentiation is a key concept in calculus that involves finding the rate at which a function changes at any given point. It is the mathematical process of deriving a function's derivative and is crucial for understanding how functions behave.

• For a function \( f(s) = \tan(s) \), the derivative \( f'(s) = \sec^2(s) \) is obtained.
• The differentiation tells us how rapidly the tangent function changes as the angle \(s\) changes.

In the context of inverse functions, differentiation helps us apply the inverse function derivative rule. Recognizing when and how to differentiate functions like trigonometric ones enables us to tackle a wide range of mathematical problems, aiding in analyzing real-world scenarios involving rates of change.

Inverse Trigonometric Functions

Inverse trigonometric functions allow for the determination of an angle given the value of a trigonometric function. They are the counterparts to the traditional trigonometric functions and play a crucial role in various applications of calculus and trigonometry.
A commonly encountered inverse trigonometric function is the arctangent function:

• The inverse of \( \tan(s) \) is \( \tan^{-1}(y) \), also known as the arctan function.
• This implies that if \( y = \tan(s) \), then \( s = \tan^{-1}(y) \).

These functions are particularly important when applying the inverse function derivative rule. Understanding these concepts allows the translation of problems that are originally posed in one function form into their inverse forms, where differentiation rules can be applied effectively.

Secant Function

The secant function, represented as \( \sec(s) \), is a vital trigonometric function often used in advanced mathematics. It is defined as the reciprocal of the cosine function:

• \( \sec(s) = \frac{1}{\cos(s)} \)
• Therefore, \( \sec^2(s) = \frac{1}{\cos^2(s)} \)

The secant function frequently appears in differentiation problems, notably when dealing with derivatives of trigonometric and inverse trigonometric functions. In our exercise, \( \sec^2(s) \) plays a critical role in finding the inverse derivative.
Its square, \( \sec^2(s) \), is key to calculating the derivative of the tangent function \( f'(s) = \sec^2(s) \). Hence, understanding this function's properties can greatly aid in applying the inverse function derivative rule and comprehending complex mathematical problems involving trigonometric functions.