If `y=(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))` then `(dy)/(dx)` at `x=0` is

Explanation

To find \(\frac{dy}{dx}\) at \(x=0\) for the function

\[ y = (1+x)(1+x^2)(1+x^4)\ldots(1+x^{2^n}), \]

we will use the product rule of differentiation.

### Step 1: Differentiate using the product rule

The product rule states that if \(y = f_1 f_2 f_3 \ldots f_n\), then

\[ \frac{dy}{dx} = f_1' f_2 f_3 \ldots f_n + f_1 f_2' f_3 \ldots f_n + f_1 f_2 f_3' \ldots f_n + \ldots + f_1 f_2 f_3 \ldots f_n'. \]

In our case, we have:

- \(f_1 = 1+x\)
- \(f_2 = 1+x^2\)
- \(f_3 = 1+x^4\)
- \(\ldots\)
- \(f_{n+1} = 1+x^{2^n}\)

### Step 2: Apply the product rule

We will differentiate each term one by one:

1. Differentiate \(f_1 = 1+x\):
\[ f_1' = 1. \]
The contribution to \(\frac{dy}{dx}\) is:
\[ 1 \cdot (1+x^2)(1+x^4)\ldots(1+x^{2^n}). \]

2. Differentiate \(f_2 = 1+x^2\):
\[ f_2' = 2x. \]
The contribution to \(\frac{dy}{dx}\) is:
\[ (1+x) \cdot 2x \cdot (1+x^4)\ldots(1+x^{2^n}). \]

3. Differentiate \(f_3 = 1+x^4\):
\[ f_3' = 4x^3. \]
The contribution to \(\frac{dy}{dx}\) is:
\[ (1+x)(1+x^2) \cdot 4x^3 \cdot (1+x^4)\ldots(1+x^{2^n}). \]

4. Continue this process until \(f_{n+1} = 1+x^{2^n}\):
\[ f_{n+1}' = 2^n x^{2^n - 1}. \]
The contribution to \(\frac{dy}{dx}\) is:
\[ (1+x)(1+x^2)(1+x^4)\ldots(1+x^{2^{n-1}}) \cdot 2^n x^{2^n - 1}. \]

### Step 3: Evaluate at \(x = 0\)

Now, we need to evaluate \(\frac{dy}{dx}\) at \(x=0\):

1. For the first term:
\[ (1+x^2)(1+x^4)\ldots(1+x^{2^n}) \text{ at } x=0 \text{ gives } 1. \]

2. For the second term:
\[ (1+x) \cdot 2x \cdot (1+x^4)\ldots(1+x^{2^n}) \text{ at } x=0 \text{ gives } 0. \]

3. For the third term:
\[ (1+x)(1+x^2) \cdot 4x^3 \cdot (1+x^4)\ldots(1+x^{2^n}) \text{ at } x=0 \text{ gives } 0. \]

4. Continuing this way, all terms involving \(x\) will vanish at \(x=0\), except for the first term.

Thus, we find that:

\[ \frac{dy}{dx} \bigg|_{x=0} = 1. \]

### Final Answer

The value of \(\frac{dy}{dx}\) at \(x=0\) is

\[ \boxed{1}. \]