If `y = 1 +(1)/(x) +(1)/(x^(2)) + (1)/(x^(3)) + ……..+oo` with `|x| gt 1` then `(dy)/(dx)` is .

Options

A. `(x^(2))/(y^(2))`
B. `x^(2)y^(2)`
C. `(y^(2))/(x^(2))`
D. `-(y^(2))/(x^(2))`

Explanation

To solve the problem, we need to find the derivative \( \frac{dy}{dx} \) for the given series:

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

### Step 1: Identify the series as a geometric series
The series can be recognized as an infinite geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = \frac{1}{x} \)

### Step 2: Use the formula for the sum of an infinite geometric series
The sum of an infinite geometric series is given by the formula:

\[ S = \frac{a}{1 - r} \]

For our series:

\[ y = \frac{1}{1 - \frac{1}{x}} \quad \text{(since } |x| > 1\text{)} \]

### Step 3: Simplify the expression for \( y \)
Substituting the values of \( a \) and \( r \):

\[ y = \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1} \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now, we need to differentiate \( y \):

\[ y = \frac{x}{x - 1} \]

Using the quotient rule for differentiation, where \( u = x \) and \( v = x - 1 \):

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = 1 \)

Substituting into the quotient rule:

\[ \frac{dy}{dx} = \frac{(x - 1)(1) - (x)(1)}{(x - 1)^2} = \frac{x - 1 - x}{(x - 1)^2} = \frac{-1}{(x - 1)^2} \]

### Step 5: Final expression for \( \frac{dy}{dx} \)
Thus, we have:

\[ \frac{dy}{dx} = -\frac{1}{(x - 1)^2} \]

### Step 6: Express \( \frac{dy}{dx} \) in terms of \( y \)
From our earlier expression for \( y \):

\[ y = \frac{x}{x - 1} \implies x - 1 = \frac{x}{y} \implies x = y(x - 1) + 1 \]

Substituting \( x - 1 = \frac{x}{y} \) into \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = -\frac{1}{\left(\frac{x}{y}\right)^2} = -\frac{y^2}{x^2} \]

### Conclusion
The final answer for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = -\frac{y^2}{x^2} \]