Problem 23 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} f(x) &=1, \quad x<0 \\ &=1+\sin x, \quad 0 \leq x<\frac{\pi}{2}, \end{aligned} $$ $$ \text { find } f^{\prime}(0) \text { . } $$

Step-by-Step Solution

Answer

\(f^{\prime}(0) = 1\)

Step 1: Analyze the function around x = 0

To find the derivative of the function at x = 0, we need to focus on the given function rules around that point. Since the second function rule applies when \(0 \leq x < \frac{\pi}{2}\), we can use this function rule to find the derivative at x=0. So, we now focus on this part: \(f(x) = 1 + \sin x, \quad 0 \leq x < \frac{\pi}{2}\)

Step 2: Differentiate the chosen function rule with respect to x

To find \(f'(x)\), we differentiate the function with respect to x: \(f'(x) = \frac{d}{dx} (1 + \sin x)\) Using the rules of differentiation, we get: \(f'(x) = \frac{d}{dx} (1) + \frac{d}{dx} (\sin x)\) \(f'(x) = 0 + \cos x\) So, \(f'(x) = \cos x\)

Step 3: Find the derivative at x = 0

Now that we have the derivative of the function, we can find the derivative at x = 0: \(f'(0) = \cos (0)\) Recall that \(\cos (0) = 1\), so we have: \(f'(0) = 1\) Therefore, \(f^{\prime}(0) = 1\).

Key Concepts

Piecewise Functions

A piecewise function is a function defined by different expressions based on the input value. In simpler terms, it's like having different rules for different parts of the number line. They are very useful in mathematical modeling because they can describe situations that change under different conditions.
For example, the function in our exercise is defined as follows:

• For values less than 0, the function is constant, specifically 1.
• For values from 0 to slightly less than \(\frac{\pi}{2}\), the function is \(1 + \sin x\).

This type of function is called 'piecewise' because it has distinct "pieces" or parts, each governed by its own rule. In our example, the transition happens at \(x = 0\).
When dealing with piecewise functions in calculus, especially when computing derivatives, it is crucial to analyze each segment separately. Then, check if we have smooth transitions at the boundaries. This requires using the rule which applies to the segment where the point of interest (in our case, \(x=0\)) lies.

Derivative

The derivative of a function at a point gives us the rate at which the function's value changes as its input changes. It is the mathematical way of determining the slope of a function at any given point. For the function from our example, characterized by segments, understanding how to handle derivatives becomes essential.
In technical terms, the derivative \(f'(x)\) of a function \(f(x)\) describes how \(f(x)\) changes as \(x\) changes, and it's mathematically expressed using differentiation. Differential calculus provides us with rules and techniques to find this derivative efficiently.

• For constants, \(c\), the derivative is 0, because constants do not change.
• For \(\sin x\), a trigonometric function, the derivative is \(\cos x\).

In our example, we calculated the derivative of the piece containing \(\sin x\). This requires applying the rules of differentiation to each part within its domain. Finally, this derivative tells us how the function behaves in the region around the specific point, \(x = 0\), where we need to find \(f'(0)\).

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, especially in calculus. They involve the relationships of angles and sides in triangles and are crucial for modeling periodic phenomena in physics and engineering.
In our function, \(\sin x\) and \(\cos x\) are the primary trigonometric functions at play. Understanding their properties and derivatives is key to solving many calculus problems.

• \(\sin x\) is a wave-like function that oscillates between -1 and 1 over each full cycle of \(2\pi\).
• The derivative of \(\sin x\) is \(\cos x\), which also oscillates but starts at 1 when \(x = 0\).
• \(\cos x\), like \(\sin x\), completes one full cycle over \(2\pi\) and oscillates between -1 and 1.

In Piecewise Functions, it's critical to remember these derivatives for functions like \(\sin x\), as seen in our exercise. Knowing that \(\cos(0) = 1\) allows us to determine the derivative of the piecewise function at the transition point \(x = 0\). This understanding helps us find smoother transitions and accurate rates of change for each piece of the function.