Problem 111 Find $f^{\prime}(x)$, where \(... [FREE SOLUTION]

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differential calculus booster with problems and solutions for iit jee main and advanced rejaul-maksh

Rejaul Makshud

$Math Studyset 91影视 Explanations$ Math

2019 Edition

Chapter 2: Problem 111

Find $f^{\prime}(x)$, where $f(x)=\cos ^{-1}(\cos x)$ and $-\pi \leq x \leq 2 \pi$

Short Answer

Expert verified

Therefore, the derivative $ f^{\prime}(x) $ of the function $ f(x) = \cos^{-1}(\cos x) $ in the interval $ -\pi \leq x \leq 2 \pi $ is equal to -1 for $ x $ in the intervals $ \[-\pi, 0] $ and $ \[\pi, 2\pi] $, and equal to 1 for $ x $ in the interval $ \[0, \pi] $.

Step by step solution

Observe behavior of function within given interval

First, note that the cosine function repeats its values in the interval $ -\pi, 2\pi $. Specifically, the output of the function $ \cos x $ for $ x = - \pi $, $ x = 0 $, and $ x = 2\pi $ is the same (all equal to 1), and the output for $ x = \pi $ is -1. Thus, the output of $ \cos^{-1}(\cos x) $ for these values of $ x $ is also the same. This indicates that the function $ f(x) = \cos^{-1}(\cos x) $ has a different behavior in $ \[-\pi, 0] $, $ \[0, \pi] $, and $ \[\pi, 2\pi] $.

Find the derivative of $ f(x) $ in each interval

In the interval $ \[-\pi, 0] $, $ f(x) = \cos^{-1}(\cos x) = -x $. The derivative of $ -x $ is $ -1 $. In the interval $ \[0, \pi] $, $ f(x) = \cos^{-1}(\cos x) = x $. The derivative of $ x $ is $ 1 $. In the interval $ \[\pi, 2\pi] $, $ f(x) = \cos^{-1}(\cos x) = 2\pi - x $. The derivative of $ 2\pi - x $ is $ -1 $. In summary, the derivative of $ f(x) $ is -1 in the intervals $ \[-\pi, 0] $ and $ \[\pi, 2\pi] $ and 1 in the interval $ \[0, \pi] $.

Compile results

Therefore, the derivative $ f^{\prime}(x) $ of the function $ f(x) = \cos^{-1}(\cos x) $ in the interval $ -\pi \leq x \leq 2 \pi $ is equal to -1 for $ x $ in the intervals $ \[-\pi, 0] $ and $ \[\pi, 2\pi] $, and is equal to 1 for $ x $ in the interval $ \[0, \pi] $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions

Trigonometric functions are fundamental in the study of mathematics, especially when dealing with angles and geometric shapes. Among these functions, cosine, denoted as cos, plays a critical role. It is defined as the ratio of the adjacent side to the hypotenuse in a right-angled triangle but also has applications in modeling periodic phenomena like waves.

When it comes to calculus, especially for exams like the IIT JEE, it's important to understand how these functions behave over specific intervals. As demonstrated in the exercise, even with inverse cosine, denoted as cos^-1, the intervals play a key role in determining the derivative. Recognizing the periodic nature of trigonometric functions and their corresponding inverse functions is crucial for calculus problems.

Derivative Concepts

The derivative of a function is a cornerstone concept in calculus, capturing the idea of instantaneous rate of change or slope of the function at a point. It's found by taking the limit as the change in the variable approaches zero. In practical terms, finding the derivative, symbolized as f' or dy/dx, allows us to determine how a function is changing at every point along its curve.

In the context of the exercise provided, understanding how to differentiate an inverse trigonometric function like inverse cosine is key. Derivatives of inverse trigonometric functions often involve applying implicit differentiation rules due to their complexity. However, in some cases, as with the given intervals, the behavior simplifies significantly, leading to elementary derivatives like -1 or 1.

Calculus for IIT JEE

Calculus is a subject with high importance in competitive exams like the Indian Institutes of Technology Joint Entrance Examination (IIT JEE). It tests students on concepts of differentiation and integration, among others. A strong grasp of calculus concepts, including the behavior of functions, the process of differentiation, and the application of these in various contexts, is required for success.

The exercise we've discussed is typical of the nuanced questions that appear on the IIT JEE. Such problems demand not only a mechanical application of derivative rules but also an understanding of function behaviors over different intervals. Questions often involve multi-layered thinking, where students are tested on their ability to recognize patterns and apply calculus concepts adeptly in solving complex problems.

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91影视

Find \(f^{\prime}(x)\), where \(f(x)=\cos ^{-1}(\cos x)\) and \(-\pi \leq x \leq 2 \pi\)

Short Answer

Step by step solution

Observe behavior of function within given interval

Find the derivative of \( f(x) \) in each interval

Compile results

Key Concepts

Trigonometric Functions

Derivative Concepts

Calculus for IIT JEE

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