Problem 273 - Calculus Volume 1

Question

Find \(\left(f^{-1}\right)^{\prime}(a)\). $$ f(x)=\tan x+3 x^{2}, a=0 $$

Step-by-Step Solution

Answer

\(\left(f^{-1}\right)'(0) = 1\)

Step 1: Understanding the Problem

We need to find the derivative of the inverse function of \( f(x) = \tan x + 3x^2 \) evaluated at \( a = 0 \). This means we want \( \left(f^{-1}\right)'(0) \).

Step 2: Apply the Derivative Formula for Inverse Functions

The derivative of an inverse function at a point is given by the formula: \[ \left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))} \] We will use this formula to find \( \left(f^{-1}\right)'(0) \). First, we have to find \( f^{-1}(0) \), the \( x \) value such that \( f(x) = 0 \).

Step 3: Solving for \( f(x) = 0 \)

Solve \( f(x) = \tan x + 3x^2 = 0 \) to find \( x \) such that \( f(x) = 0 \). As \( a = 0 \), \( f^{-1}(0) \) is the \( x \) value we need to find. Set up the equation: \( \tan x + 3x^2 = 0 \). For small \( x \), \( \tan x \approx x \), leading to the approximation \( x + 3x^2 \approx 0 \). Solving for small \( x \), we get \( x = 0 \) as a solution.

Step 4: Calculate \( f'(x) \)

Find the derivative \( f'(x) \). \( f(x) = \tan x + 3x^2 \), so \( f'(x) = \sec^2 x + 6x \).

Step 5: Evaluate \( f'(x) \) at \( x = 0 \)

Evaluate the derivative \( f'(x) \) at \( x = 0 \). Substitute \( x = 0 \) into the derivative: \( f'(0) = \sec^2(0) + 6 \times 0 = 1 + 0 = 1 \).

Step 6: Calculate \( \left(f^{-1}\right)'(0) \)

Using the inverse function derivative formula from Step 2, \( \left(f^{-1}\right)'(0) = \frac{1}{f'(0)} = \frac{1}{1} = 1 \).

Key Concepts

Derivative of Inverse Functions

Derivatives of inverse functions can seem a bit daunting at first, but they're simplified through a handy formula. The key idea is to find the derivative of the inverse function by working with the derivative of the function itself. Essentially, if you have a function \( f(x) \), and you're looking to find the derivative of its inverse, the formula you use is:\[ \left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))} \]This formula tells us that the derivative of the inverse function at \( a \) is the reciprocal of the derivative of \( f \) at the point \( f^{-1}(a) \). Here's a basic rundown of how this operates:

• First, you find \( f^{-1}(a) \), which is the \( x \) such that \( f(x) = a \).
• Then, you find \( f'(x) \), which is the derivative of \( f \).
• Finally, you evaluate \( f'(x) \) at \( x = f^{-1}(a) \), and take the reciprocal to find \( \left(f^{-1}\right)'(a) \).

By understanding this procedure, you can readily apply the formula and tackle complex calculus questions with confidence.

Trigonometric Functions

Trigonometric functions like \( \tan x \) play a significant role when working with derivatives. These functions often appear in calculus problems due to their unique properties and derivatives. For instance, the derivative of \( \tan x \) is \( \sec^2 x \), which is a vital detail in derivative calculations.When you're solving problems involving trigonometric functions, it's useful to remember their derivatives and basic identities:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

It's also helpful to approximate trigonometric functions for small angles. For small \( x \), \( \tan x \approx x \). These approximations can simplify calculus problems, especially when solving equations involving trigonometric components alongside other terms.

Calculus Problems

Solving calculus problems often involves multiple steps and concepts, bringing together derivatives, inverses, and trigonometric functions. Let's break down a typical approach:1. **Understanding the Problem**:
It can involve finding a derivative, such as an inverse derivative, or solving an equation.2. **Using Formulas**:
Applying the right derivative formula depending on whether you're dealing with direct or inverse functions.3. **Setting up Equations**:
Sometimes, you'll form equations that require solving for unknowns, such as finding \( x \) when \( f(x) = a \).Consider the problem from above, which asks for \( \left(f^{-1}\right)'(0) \). Solving these kinds of problems involves carefully considering both the trigonometric function, such as \( \tan x \), and any polynomial terms like \( 3x^2 \). You calculate derivatives with respect to each part, combine results, and evaluate at required points. Through practice and understanding, solving complex calculus problems becomes manageable and straightforward.