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

Options

A. `-1`
B. 0
C. 1
D. None

Answer

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

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

we will follow these steps:

### Step 1: Rewrite the function
We can express \(y\) as a product of terms:

\[ y = \prod_{k=0}^{n} (1 + x^{2^k}). \]

### Step 2: Differentiate using the product rule
To differentiate \(y\), we will use the product rule. The derivative of a product of functions is given by:

\[ \frac{dy}{dx} = \sum_{k=0}^{n} \left( \prod_{j \neq k} (1 + x^{2^j}) \cdot \frac{d}{dx}(1 + x^{2^k}) \right). \]

The derivative of \(1 + x^{2^k}\) is \(2^k x^{2^k - 1}\). Thus, we have:

\[ \frac{dy}{dx} = \sum_{k=0}^{n} \left( \prod_{j \neq k} (1 + x^{2^j}) \cdot 2^k x^{2^k - 1} \right). \]

### Step 3: Evaluate at \(x = 0\)
Now, we will evaluate \(\frac{dy}{dx}\) at \(x = 0\):

1. When \(x = 0\), each term \(1 + x^{2^j}\) becomes \(1\).
2. The derivative term \(2^k x^{2^k - 1}\) becomes \(0\) for all \(k\) except when \(k = 0\) (since \(x^{2^k - 1} = 0\) for \(k > 0\)).
3. For \(k = 0\), we have \(2^0 x^{2^0 - 1} = 1\).

Thus, at \(x = 0\):

\[ \frac{dy}{dx} = \prod_{j \neq 0} (1 + 0^{2^j}) \cdot 1 = 1 \cdot 1 = 1. \]

### Final Result
Therefore, the value of \(\frac{dy}{dx}\) at \(x = 0\) is:

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

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