Problem 22 - A Textbook of Differential Calculus

Question

If \(f(x)=|\cos x-\sin x|\) then \(f^{\prime}\left(\frac{\pi}{4}\right)\) is equal to (a) \(\sqrt{2}\) (b) \(-\sqrt{2}\) (c) 0 (d) None of these

Step-by-Step Solution

Answer

The derivative \(f^{\prime}(\frac{\pi}{4})\) is 0. Therefore, the answer is (c) 0.

Step 1: Understand the Absolute Function

First, recognize that the function given is a composition involving an absolute value: \[ f(x) = |\cos x - \sin x|. \]This means for certain intervals of \(x\), the function \(\cos x - \sin x\) changes sign. So, \(f(x)\) can be written piecewise as either \(\cos x - \sin x\) or \(-(\cos x - \sin x)\) depending on whether \(\cos x - \sin x\) is positive or negative.

Step 2: Determine Critical Points for Sign Changes

To determine when \(\cos x - \sin x\) changes sign, set \(\cos x - \sin x = 0\). This gives us:\[ \cos x = \sin x. \]This equation holds true when \(x = \frac{\pi}{4} + n\pi\) for integer \(n\). Thus, at these points, the expression \(\cos x - \sin x\) crosses zero, indicating possible changes in sign.

Step 3: Evaluate Behavior at x = π/4

Evaluate \(\cos x - \sin x\) at \(x = \frac{\pi}{4} \) to determine its sign. We have:\[\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \]Thus:\[ \cos \frac{\pi}{4} - \sin \frac{\pi}{4} = 0. \]The changeover point is exactly at \(x = \frac{\pi}{4}\), so the behavior of the function needs careful consideration around this point.

Step 4: Express f(x) Near x = π/4

For \(x\) just less than \(\frac{\pi}{4}\), \(\sin x\) is slightly greater than \(\cos x\) (leading to \(\cos x - \sin x < 0\) ) and for \(x\) just greater, \(\cos x\) is slightly greater (leading to \(\cos x - \sin x > 0\) ). Hence, the function \(f(x)\) switches from:\[ f(x) = - (\cos x - \sin x) \quad \text{to} \quad f(x) = \cos x - \sin x.\]

Step 5: Differentiate f(x) for x ≠ π/4

The derivative of \(f(x)\) is piecewise, depending on \(x\) compared to \(\frac{\pi}{4}\):\[ \text{If } x < \frac{\pi}{4}, \, f(x) = - (\cos x - \sin x), \text{ so } f'(x) = \sin x + \cos x. \]\[ \text{If } x > \frac{\pi}{4}, \, f(x) = \cos x - \sin x, \text{ so } f'(x) = -\sin x - \cos x. \]

Step 6: Evaluate the Limit of the Derivative at x = π/4

We need to evaluate the derivative approaching \(\frac{\pi}{4}\) from both sides:- From below, \(f'(x) = \sin x + \cos x\).- From above, \(f'(x) = -\sin x - \cos x\).Both limits approach 0 as \(x\) approaches \(\frac{\pi}{4}\), as both \(\sin\frac{\pi}{4}\) and \(\cos\frac{\pi}{4}\) equal \(\frac{\sqrt{2}}{2}\). Thus, the derivative \(f'(\frac{\pi}{4}) = 0\).

Key Concepts

Absolute Value Function

An absolute value function represents the magnitude of a value, disregarding its sign. In simpler terms, it's like measuring distance without worrying about the direction. The function given in the exercise is an absolute value function: \[ f(x) = |\cos x - \sin x|. \]This equation ensures that regardless of whether \(\cos x - \sin x\) is positive or negative, the value of \(f(x)\) will remain non-negative.

• This absolute value can create a 'V' shaped graph, as it reflects all negative values to positive by mirroring them over the x-axis.
• Critical points of interest often occur where the inside equation, \(\cos x - \sin x\), equals zero since those are the points of reflection or change in behavior of the original function.

Piecewise Function

Piecewise functions are defined by different expressions in different parts of their domain. They effectively combine multiple subfunctions under a single umbrella function, each applicable within specified conditions or intervals. For \(f(x) = |\cos x - \sin x|\), the function can be expressed in a piecewise manner:

• If \(\cos x - \sin x \geq 0\), then \(f(x) = \cos x - \sin x\).
• If \(\cos x - \sin x < 0\), then \(f(x) = - (\cos x - \sin x)\).

This technique allows us to handle the absolute value function by covering all possible values of \(x\) and providing a complete description of the function, taking into account both positive and negative scenarios on \(\cos x - \sin x\).

Derivative

In calculus, the derivative represents the rate at which a function is changing at any point. It's like measuring how fast you're driving by looking at your speedometer at any instant. For the absolute value function, \(f(x) = |\cos x - \sin x|\), the derivative must be considered carefully due to the piecewise nature of the function.

• For the subdomain where \(x < \frac{\pi}{4}\), \(f(x) = - (\cos x - \sin x)\), which leads to the derivative:\[ f'(x) = \sin x + \cos x. \]
• For the subdomain where \(x > \frac{\pi}{4}\), \(f(x) = \cos x - \sin x\), resulting in the derivative:\[ f'(x) = -\sin x - \cos x. \]

When evaluating at \(x = \frac{\pi}{4}\), careful consideration of both one-sided limits reveals that the derivative approaches 0.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in studying wave patterns and periodic phenomena. They oscillate between -1 and 1 and help manage periodic changes over time. The function \(\cos x - \sin x\) illustrates how combining these functions can create complex behavior:

• At \(x = \frac{\pi}{4}\), both \(\cos \frac{\pi}{4}\) and \(\sin \frac{\pi}{4}\) equal \(\frac{\sqrt{2}}{2}\), perfectly balancing each other.
• Elsewhere, depending on the interval relative to \(\frac{\pi}{4}\), one of the functions will dominate slightly, causing \(\cos x - \sin x\) to switch its sign.

Thus, understanding the nature of trigonometric functions is essential for evaluating function behavior within specified intervals and managing the transition between different states in a piecewise format.