Problem 227 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\text { Check the differentiability of the function \& find } f^{\prime}(x) \text { . }\\\ &\begin{aligned} f(x) &=\sin x, & x \geq 0 \\ &=e^{x}-1, & x<0 \end{aligned} \end{aligned} $$

Step-by-Step Solution

Answer

The given function is differentiable for all x, and its derivative is f'(x) = \(e^x\) for x < 0 and f'(x) = \(cos x\) for x ≥ 0.

Step 1: Determine differentiability at x = 0

First, we need to check if the function, f(x), is differentiable at x = 0. To do this, we will find the left-hand derivative (LHD) and right-hand derivative (RHD) at x = 0 and check if they are equal.

Step 2: Calculate the left-hand derivative (LHD) at x = 0

For x < 0, the function is defined as f(x) = e^x - 1. To find the LHD at x = 0, we differentiate this part of the function with respect to x: LHD = \(\lim_{h\to 0} \frac{f(0-h) - f(0)}{-h}\) Where f(x) = e^x - 1. Therefore, LHD becomes: LHD = \(\lim_{h\to 0} \frac{e^{-h} - 1}{-h}\)

Step 3: Calculate the right-hand derivative (RHD) at x = 0

For x ≥ 0, the function is defined as f(x) = sin x. To find the RHD at x = 0, we differentiate this part of the function with respect to x: RHD = \(\lim_{h\to 0} \frac{f(h) - f(0)}{h}\) Where f(x) = sin x. Therefore, RHD becomes: RHD = \(\lim_{h\to 0} \frac{\sin h}{h}\)

Step 4: Check if LHD = RHD

To check if the function is differentiable at x = 0, we need to verify if LHD = RHD. Using the given limits, we have: LHD = \(\lim_{h\to 0} \frac{e^{-h} - 1}{-h} = 1\) RHD = \(\lim_{h\to 0} \frac{\sin h}{h} = 1\) As LHD = RHD, the function is differentiable at x = 0.

Step 5: Find the derivative of the function for x ≠ 0

Now that we know the function is differentiable at x = 0, we need to find its derivative for x ≠ 0. For x < 0, the function is defined as f(x) = e^x - 1, and for x ≥ 0, the function is defined as f(x) = sin x. Derivative of f(x) when x < 0: f'(x) = \(\frac{d}{dx}\) (e^x - 1) = e^x Derivative of f(x) when x ≥ 0: f'(x) = \(\frac{d}{dx}\) (sin x) = cos x So, the derivative of the function f(x) is: f'(x) = \( \begin{cases} e^x, & \text{if } x < 0 \\ cos x, & \text{if } x \ge 0 \end{cases} \) Hence, the given function is differentiable for all x, and its derivative is f'(x) = e^x for x < 0 and f'(x) = cos x for x ≥ 0.

Key Concepts

Left-Hand Derivative

The left-hand derivative (LHD) of a function at a particular point provides us insight into how the function behaves as we approach that point from the left side. If we have a function, like the one in our exercise, defined in a piecewise manner, we need a clear understanding of how it changes as we pivot our approach direction, particularly from the left.
The left-hand derivative is calculated using the following formula:

• LHD = \(\lim_{h\to 0} \frac{f(a-h) - f(a)}{-h}\)
• Here, \(a\) represents the point where we're assessing differentiability.
• The term \(f(a-h)\) indicates that we're looking at values slightly to the left of \(a\).

For example, in our piecewise function, where \(f(x) = e^x - 1\) for \(x < 0\), the left-hand derivative at \(x = 0\) is:

• LHD = \(\lim_{h\to 0} \frac{e^{-h} - 1}{-h}\)

Calculating this limit gives us a numerical value, helping us confirm the behavior of the function as we approach zero from the negative side.

Right-Hand Derivative

On the flip side, the right-hand derivative (RHD) examines how a function behaves as we approach a point from the right, or positive side. It's especially crucial for functions defined in parts, sewn together via conditions at key points.
Calculating the right-hand derivative involves the following process:

• RHD = \(\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}\)
• Here, \(a\) is the same point we are focusing on, with \(a+h\) indicating the values just to the right of \(a\).

For our piecewise example, where \(f(x) = \sin x\) for \(x \geq 0\), the right-hand derivative at \(x = 0\) manifests as:

• RHD = \(\lim_{h\to 0} \frac{\sin h}{h}\)

Solving this limit helps us uncover the function's tendency as one approaches zero from the positive side. If both LHD and RHD are equal, the function is deemed differentiable at that point.

Piecewise Function

A piecewise function consists of different functions applied to various parts of its domain. Such functions are defined by multiple sub-functions, each governing a segment of the main function's path. This property makes piecewise functions extremely flexible, yet sometimes complex, when analyzing differentiability or continuity.

This kind of function utilizes separate formulas with specific domain allocations:

• Example: \(f(x) = \sin x\) for \(x \geq 0\), and \(f(x) = e^x - 1\) for \(x < 0\).
• The two expressions describe behavior in different segments, neatly joined at intersections like \(x = 0\).

Examining these connections requires us to calculate both left and right-hand derivatives and ensure they match. Only then can we say if the function is smoothly joined and differentiable at that criteria point. By understanding this fundamental aspect, you demystify differentiability at mix-matched junctures. A key takeaway: check if the formulas meet seamlessly at the boundaries.