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

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

we will first differentiate \(y\) with respect to \(x\) and then evaluate the derivative at \(x=0\).

### Step 1: Write down the function
The function is given as:

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

### Step 2: Differentiate using the product rule
To differentiate \(y\), we will use the product rule. The product rule states that if \(y = u_1 u_2 u_3 \cdots u_k\), then

\[ \frac{dy}{dx} = u_1' u_2 u_3 \cdots u_k + u_1 u_2' u_3 \cdots u_k + \cdots + u_1 u_2 u_3 \cdots u_k'. \]

In our case, we have \(k = n+1\) terms: \(u_1 = 1+x\), \(u_2 = 1+x^2\), ..., \(u_{n+1} = 1+x^{2^n}\).

### Step 3: Differentiate each term
The derivative of each term is:

- \(u_1' = 1\)
- \(u_2' = 2x\)
- \(u_3' = 4x^3\)
- ...
- \(u_{n+1}' = 2^n x^{2^n - 1}\)

### Step 4: Apply the product rule
Using the product rule, we can express \(\frac{dy}{dx}\) as:

\[ \frac{dy}{dx} = (1)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) + (1+x)(2x)(1+x^4)\cdots(1+x^{2^n}) + (1+x)(1+x^2)(4x^3)(1+x^4)\cdots(1+x^{2^n}) + \ldots \]

### Step 5: Evaluate at \(x=0\)
Now we will evaluate \(\frac{dy}{dx}\) at \(x=0\):

1. For the first term:
\[ (1)(1+0^2)(1+0^4)\cdots(1+0^{2^n}) = 1. \]

2. For the second term:
\[ (1+0)(2 \cdot 0)(1+0^4)\cdots(1+0^{2^n}) = 0. \]

3. For the third term:
\[ (1+0)(1+0^2)(4 \cdot 0^3)(1+0^4)\cdots(1+0^{2^n}) = 0. \]

Continuing this way, all terms involving \(x\) will vanish, and we will only have the first term contributing to the derivative.

### Final Result
Thus, we find that:

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