if `f (x) = lim _(x to oo) (prod_(i=l)^(n ) cos ((x )/(2 ^(l))))` then f '(x) is equal to:

Options

A. `(sin x)/(x)`
B. `(x)/(sin x)`
C. `(x cos - sin x)/(x ^(2))`
D. `(sin x-x cos x)/(sin ^(2)x)`

Answer

To solve the problem, we need to find the derivative \( f'(x) \) of the function defined as:

\[ f(x) = \lim_{x \to \infty} \prod_{i=1}^{n} \cos\left(\frac{x}{2^i}\right) \]

### Step-by-Step Solution:

1. **Understanding the Product**:
The function \( f(x) \) is defined as the limit of the product of cosine functions. We can rewrite it as:
\[ f(x) = \lim_{x \to \infty} \left( \cos\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{4}\right) \cdot \cos\left(\frac{x}{8}\right) \cdots \cos\left(\frac{x}{2^n}\right) \right) \]

2. **Using the Identity**:
We can use the identity \( 2 \sin A \cos A = \sin(2A) \) to simplify the product. We can express the product in terms of sine:
\[ f(x) = \lim_{x \to \infty} \frac{\sin\left(\frac{x}{2^{n-1}}\right)}{2^{n-1}} \cdot \frac{1}{\sin\left(\frac{x}{2^n}\right)} \]

3. **Taking the Limit**:
As \( x \to \infty \), the sine function oscillates between -1 and 1. Thus, the limit of the product will depend on the behavior of the sine terms:
\[ f(x) = \lim_{x \to \infty} \frac{\sin\left(\frac{x}{2^{n-1}}\right)}{2^{n-1} \sin\left(\frac{x}{2^n}\right)} \]

4. **Finding the Derivative**:
To find \( f'(x) \), we will apply the product and chain rule. We can express \( f(x) \) in terms of two functions \( u(x) \) and \( v(x) \):
\[ u(x) = \sin\left(\frac{x}{2^{n-1}}\right), \quad v(x) = \sin\left(\frac{x}{2^n}\right) \]
Then,
\[ f(x) = \frac{u(x)}{2^{n-1} v(x)} \]

5. **Applying the Quotient Rule**:
The derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2 \cdot 2^{n-1}} \]

6. **Calculating \( u'(x) \) and \( v'(x) \)**:
Using the chain rule:
\[ u'(x) = \frac{1}{2^{n-1}} \cos\left(\frac{x}{2^{n-1}}\right), \quad v'(x) = \frac{1}{2^n} \cos\left(\frac{x}{2^n}\right) \]

7. **Substituting Back**:
Substitute \( u'(x) \) and \( v'(x) \) back into the derivative formula:
\[ f'(x) = \frac{\frac{1}{2^{n-1}} \cos\left(\frac{x}{2^{n-1}}\right) \sin\left(\frac{x}{2^n}\right) - \sin\left(\frac{x}{2^{n-1}}\right) \frac{1}{2^n} \cos\left(\frac{x}{2^n}\right)}{(\sin\left(\frac{x}{2^n}\right))^2 \cdot 2^{n-1}} \]

### Final Result:
Thus, the derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{\frac{1}{2^{n-1}} \cos\left(\frac{x}{2^{n-1}}\right) \sin\left(\frac{x}{2^n}\right) - \sin\left(\frac{x}{2^{n-1}}\right) \frac{1}{2^n} \cos\left(\frac{x}{2^n}\right)}{(\sin\left(\frac{x}{2^n}\right))^2 \cdot 2^{n-1}} \]