Problem 16 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

The function \(f(x)=\tan ^{-1}(\sin x+\cos x)\) is an increasing function in [AIEEE-2007] (a) \((\pi / 4, \pi / 2)\) (b) \((-\pi / 2, \pi / 4)\) (c) \((0, \pi / 2)\) (d) \((-\pi / 2, \pi / 2)\)

Step-by-Step Solution

Answer

(b) \((-\pi/2, \pi/4)\)

Step 1: Understand the function

The function given is \( f(x) = \tan^{-1}(\sin x + \cos x) \). This is the inverse tangent of the sum of \( \sin x \) and \( \cos x \), and we're asked to identify where this function is increasing.

Step 2: Find the derivative

To determine where \( f(x) \) is increasing, we need to find its derivative. Use the chain rule to differentiate \( f(x) \): \( f'(x) = \frac{d}{dx}[\tan^{-1}(u)] \cdot \frac{d}{dx}(\sin x + \cos x) \) where \( u = \sin x + \cos x \).

Step 3: Differentiate the inside function

The derivative of \( u = \sin x + \cos x \) is \( u' = \cos x - \sin x \).

Step 4: Differentiate the inverse function

The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \). So the derivative of \( f(x) \) is \( f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x) \).

Step 5: Simplify the derivative

Simplify \( f'(x) \): Expand \( (\sin x + \cos x)^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x \). Since \( \sin^2 x + \cos^2 x = 1 \), \( f'(x) = \frac{\cos x - \sin x}{2\sin x \cos x + 2}. \)

Step 6: Determine where function is increasing

For \( f(x) \) to be increasing, \( f'(x) > 0 \). This means finding where \( \cos x - \sin x > 0 \). Notice that \( \cos x = \sin(x + \frac{\pi}{4}) \) makes \( \cos x - \sin x = \sqrt{2}(\cos x') \), indicating where \( x \) is between \(-\frac{\pi}{4}\) and \( \frac{\pi}{4} \).

Step 7: Evaluate the options

Check the given intervals to find where \( x \) lies between \(-\frac{\pi}{4}\) and \( \frac{\pi}{4} \). Only the interval \((-\pi/2, \pi/4)\) includes the range that makes \( f'(x) \) positive within that interval.

Key Concepts

Increasing Functions

An increasing function is one where, as the input value (usually referred to as "x") increases, the output value ("f(x)") also increases. This means a positive slope or a rising curve when you visually represent it on a graph.
For a function to be increasing over a particular interval, its derivative must be greater than zero within that range. In simpler terms, if you imagine walking along the curve from left to right, you should always find yourself walking uphill. Here's a breakdown of what's happening in this specific problem with the function \( f(x) = \tan^{-1} (\sin x + \cos x) \).

• The function is a composition of trigonometric and inverse trigonometric functions.
• To check if the function is increasing, we need to look at the derivative \( f'(x) \).
• This derivative will help us determine where the function changes direction or stays consistent in its upward climb.

The important part here is to find where \( f'(x) > 0 \). After deriving \( f'(x) \) and simplifying, we see that the result dictates the intervals where the function is ascending.

Derivatives in Calculus

Derivatives are fundamental to calculus because they provide information about the rate of change of a function with respect to a variable. Essentially, the derivative tells us how quickly or slowly a function is changing at any given point.
In our function \( f(x) = \tan^{-1} (\sin x + \cos x) \), the derivative allows us to assess whether the function is increasing or decreasing. To find the derivative, we apply the chain rule, a method used when dealing with composite functions — where one function is inside another.

• The derivative \( f'(x) = \frac{d}{dx}[\tan^{-1}(u)] \cdot \frac{d}{dx}(\sin x + \cos x) \), where \( u = \sin x + \cos x \), uses both the chain rule and known derivatives of trigonometric functions.
• The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \), while the derivative of \( \sin x + \cos x \) is \( \cos x - \sin x \).
• By multiplying these results, \( f'(x) \) is obtained, which helps us determine intervals where the function is increasing.

Understanding derivatives is crucial because it gives a mathematical explanation for changes and behaviors of functions, leading us to make informed conclusions about how the function acts in different sections of its domain.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function. Different functions require different techniques to differentiate them correctly. Let's dive into the techniques used in this exercise.

• **Chain Rule**: This technique is essential when dealing with composite functions — functions within functions. For example, in \( f(x) = \tan^{-1}(\sin x + \cos x) \), \( \tan^{-1} \) is the outer function and \( \sin x + \cos x \) the inner function.
• **Product and Sum Rules**: Although we didn't directly use these here, they apply to functions where you multiply several components or when they are added/subtracted, informing separate differentiation approaches.
• **Trigonometric Derivatives**: Knowing specific derivatives helps. The derivatives of \( \sin x \) and \( \cos x \) are \( \cos x \) and \(-\sin x \), respectively. These fundamental derivatives play a role in this exercise.

By combining these techniques, we can solve more complex differentiation problems comprehensively, such as establishing where a complex function like \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing or decreasing.