If y= ` 1+x+ (x^(2)) /(2!) +(x^(3))/( 3!) +......infty ,then (dy)/(dx) ` =

Options

A. y
B. `-y`
C. 1
D. 0

Answer

To solve the problem, we need to differentiate the given infinite series representation of \( y \):

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

This series is the Taylor series expansion for \( e^x \). However, let's derive \( \frac{dy}{dx} \) step by step.

### Step 1: Write down the function
We start with the function:

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

### Step 2: Differentiate term by term
We differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x\right) + \frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \ldots \]

### Step 3: Calculate each derivative
Calculating the derivatives of each term:

1. The derivative of \( 1 \) is \( 0 \).
2. The derivative of \( x \) is \( 1 \).
3. The derivative of \( \frac{x^2}{2!} \) is \( \frac{2x}{2!} = \frac{x}{1!} \).
4. The derivative of \( \frac{x^3}{3!} \) is \( \frac{3x^2}{3!} = \frac{x^2}{2!} \).
5. Continuing this process, the \( n^{th} \) term \( \frac{x^n}{n!} \) will differentiate to \( \frac{nx^{n-1}}{n!} = \frac{x^{n-1}}{(n-1)!} \).

### Step 4: Rewrite the series
Thus, we can rewrite the differentiated series:

\[ \frac{dy}{dx} = 0 + 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

### Step 5: Recognize the series
Notice that this series is the same as the original series for \( y \):

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

### Final Result
Thus, we conclude that:

\[ \frac{dy}{dx} = y \]