Find the derivative of the function given by :
`f(x)=(1+x)(1+x^(2))(1+x^(4)).....(1+x^(2n))` and hence, find f'(0).

Options

A. 0
B. -1
C. 1
D. 2

Answer

To find the derivative of the function \( f(x) = (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2n}) \) and subsequently \( f'(0) \), we will follow these steps:

### Step 1: Take the logarithm of the function
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \log f(x) = \log((1+x)(1+x^2)(1+x^4)\cdots(1+x^{2n})) \]

### Step 2: Use properties of logarithms
Using the property of logarithms that states \( \log(a \cdot b) = \log a + \log b \), we can expand the right-hand side:
\[ \log f(x) = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \cdots + \log(1+x^{2n}) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides:
\[ \frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)} \]
For the right-hand side, we differentiate each term:
\[ \frac{d}{dx}(\log(1+x^k)) = \frac{kx^{k-1}}{1+x^k} \]
Thus, we have:
\[ \frac{f'(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots + \frac{2n x^{2n-1}}{1+x^{2n}} \]

### Step 4: Multiply both sides by \( f(x) \)
Rearranging gives us:
\[ f'(x) = f(x) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots + \frac{2n x^{2n-1}}{1+x^{2n}} \right) \]

### Step 5: Evaluate \( f(0) \)
Next, we need to find \( f(0) \):
\[ f(0) = (1+0)(1+0)(1+0)\cdots(1+0) = 1 \cdot 1 \cdot 1 \cdots \cdot 1 = 1 \]

### Step 6: Evaluate \( f'(0) \)
Now, we substitute \( x = 0 \) into the expression for \( f'(x) \):
\[ f'(0) = f(0) \left( \frac{1}{1+0} + \frac{2 \cdot 0}{1+0^2} + \frac{4 \cdot 0^3}{1+0^4} + \cdots + \frac{2n \cdot 0^{2n-1}}{1+0^{2n}} \right) \]
This simplifies to:
\[ f'(0) = 1 \left( 1 + 0 + 0 + \cdots + 0 \right) = 1 \]

### Final Answer
Thus, the derivative of the function at \( x = 0 \) is:
\[ f'(0) = 1 \]