If ` y = 1 - x +(x^(2))/(2!) - (x^(3))/(3!) + (x^(4))/(4!) - ..., " then" (d^(2) y)/(dx^(2)) ` is equal to

Options

A. x
B. `-x`
C. `-y`
D. y

Explanation

To solve the problem, we need to find the second derivative of the function given by the series:

\[ y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \ldots \]

This series resembles the Taylor series expansion of the exponential function. Specifically, it can be recognized as the series for \( e^{-x} \).

### Step 1: Recognize the Series
The series can be rewritten as:
\[ y = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \]

This is the Taylor series for \( e^{-x} \).

### Step 2: Write the Function
Thus, we can express \( y \) as:
\[ y = e^{-x} \]

### Step 3: First Derivative
Now, we will find the first derivative of \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{-x}) \]
Using the chain rule:
\[ \frac{dy}{dx} = -e^{-x} \]

### Step 4: Second Derivative
Next, we find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-e^{-x}) \]
Again using the chain rule:
\[ \frac{d^2y}{dx^2} = -\frac{d}{dx}(e^{-x}) \]
\[ \frac{d^2y}{dx^2} = -(-e^{-x}) \]
\[ \frac{d^2y}{dx^2} = e^{-x} \]

### Step 5: Substitute Back
Since we established that \( y = e^{-x} \), we can substitute back:
\[ \frac{d^2y}{dx^2} = y \]

### Conclusion
Thus, we conclude that:
\[ \frac{d^2y}{dx^2} = e^{-x} = y \]

So, the final answer is:
\[ \frac{d^2y}{dx^2} = y \]