Problem 4 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\cos x \cos 2 x \cos 4 x \cos 8 x\), then \(f^{\prime}\left(\frac{\pi}{4}\right)\) is (A) \(-1\) (B) 2 (C) \(\sqrt{2}\) (D) None of these

Step-by-Step Solution

Answer

Option C: \( \sqrt{2} \).

Step 1: Differentiate the function

To find the derivative of the function \( f(x) = \cos x \cdot \cos 2x \cdot \cos 4x \cdot \cos 8x \), we apply the product rule for derivatives which states that the derivative of a product of functions is the first times the derivative of the second plus the second times the derivative of the first, and so on.Let \( u = \cos x \), \( v = \cos 2x \), \( w = \cos 4x \), and \( z = \cos 8x \). Therefore,\[f'(x) = \frac{d}{dx}[u \cdot v \cdot w \cdot z]\]Apply the product rule:\[f'(x) = u'vwz + uv'wz + uvw'z + uvwz'\]

Step 2: Compute individual derivatives

Calculate the derivatives of each component:- \( u' = \frac{d}{dx}(\cos x) = -\sin x \)- \( v' = \frac{d}{dx}(\cos 2x) = -2 \sin 2x \)- \( w' = \frac{d}{dx}(\cos 4x) = -4 \sin 4x \)- \( z' = \frac{d}{dx}(\cos 8x) = -8 \sin 8x \)Substitute back into the derivative equation:\[ f'(x) = -\sin x \cdot \cos 2x \cdot \cos 4x \cdot \cos 8x + \cos x \cdot (-2 \sin 2x) \cdot \cos 4x \cdot \cos 8x + \cos x \cdot \cos 2x \cdot (-4 \sin 4x) \cdot \cos 8x + \cos x \cdot \cos 2x \cdot \cos 4x \cdot (-8 \sin 8x) \]

Step 3: Evaluate at \( x = \frac{\pi}{4} \)

Substitute \( x = \frac{\pi}{4} \) into the equation:- \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)- \( \cos\left(\frac{\pi}{2}\right) = 0 \)- \( \sin\left(\frac{\pi}{2}\right) = 1 \)- \( \cos(\pi) = -1 \)- \( \sin(\pi) = 0 \)Thus the terms become:\[ \begin{align*} f'\left(\frac{\pi}{4}\right) = & -\frac{\sqrt{2}}{2} \cdot 0 \cdot \cos\left(\pi\right) \cdot -1 + \cos\left(\frac{\pi}{4}\right) \cdot -2 \cdot 1 \cdot -1 \ & + \cos\left(\frac{\pi}{4}\right) \cdot 0 \cdot \cos\left(\pi\right) \cdot -1 + \cos\left(\frac{\pi}{4}\right) \cdot 0 \cdot \cos\left(\pi\right) \cdot 1 \end{align*} \]

Step 4: Simplify term by term

Evaluating each non-zero component:- The first and third terms are zero because they contain \( \cos\left(\frac{\pi}{2}\right) = 0 \).- The second term is \( \cos\left(\frac{\pi}{4}\right) \cdot 2 \cdot 1 \cdot (-1) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \).- The fourth term is zero because it contains \( \sin(\pi) = 0 \).Thus, we obtain:\[ f'\left(\frac{\pi}{4}\right) = \sqrt{2} \].

Step 5: Conclude with the answer

The derivative of the function \( f(x) \) at \( x = \frac{\pi}{4} \) is \( \sqrt{2} \). Therefore, the correct answer is option C: \( \sqrt{2} \).

Key Concepts

Derivative

The concept of a derivative is a fundamental component in calculus. Simply put, it's the measure of how a function changes as its input changes. In more formal terms, the derivative represents the instantaneous rate of change of the function with respect to one of its variables. When dealing with derivatives, there are a few essential points to remember:

• The notation for derivative involves symbols like \( f'(x) \) or \( \frac{dy}{dx} \), which indicate the derivative of \( y \) with respect to \( x \).
• The process of finding a derivative is called differentiation.
• A derivative is often thought of as the slope of the tangent line to the graph of the function at a given point.

To use derivatives effectively, understanding these principles is crucial because they form the basis for more advanced calculus concepts.

Trigonometric Functions

Trigonometric functions are a set of functions related to angles and essential in many areas of mathematics and science. These include functions like sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)).When using trigonometric functions:

• They are periodic, meaning they repeat values in a regular interval.
• For sine and cosine, their values oscillate between -1 and 1.
• Trigonometric functions are important for representing wave-like patterns, among other things.

An example of applying this concept is calculating angles or sides of triangles in geometry. Additionally, they often appear in calculus problems, especially within derivatives, where the rate of change of these trigonometric functions can be determined by their respective derivatives like \( \frac{d}{dx}(\cos x) = -\sin x \). Understanding these derivatives can simplify dealing with problems that involve periodic or oscillating phenomena.

Product Rule

The product rule is a critical tool in differentiation when dealing with the product of two or more functions. It allows us to find the derivative of a product of functions efficiently.The product rule formula is:

• For two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \, v(x) \) is \( u'(x) \, v(x) + u(x) \, v'(x) \).
• When applied to more functions, such as \( u(x), v(x), w(x), z(x) \), the formula extends to include each function sequentially deriving its own turn while holding the others constant:
• \( u'(x) \, v(x) \, w(x) \, z(x) + u(x) \, v'(x) \, w(x) \, z(x) + u(x) \, v(x) \, w'(x) \, z(x) + u(x) \, v(x) \, w(x) \, z'(x) \).

This rule is especially helpful in complex problems where straightforward derivatives are not suitable, enabling us to manage and compute derivatives of products without transforming the original expression into a more extensive or cumbersome form.