Question

If \[ y = 1 + x + x^2 + x^3 + \dots \quad \text{and} \quad |x|<1, \text{ then } y'' = \]

• A. \( 2yy' \)
• B. \( \frac{2y}{y'} \)
• C. \( \frac{y'}{2y} \)
• D. \( 2y^2y' \)

Hint:

For infinite geometric series, recognize the closed-form formula and use basic differentiation rules to simplify expressions.

Answer 1

The Correct Option is A

Step 1: Recognizing the sum of an infinite geometric series
The given function \( y \) represents the sum of an infinite geometric series: \[ y = \sum_{n=0}^{\infty} x^n. \] Using the formula for the sum of an infinite geometric series: \[ y = \frac{1}{1 - x}, \quad \text{for } |x|<1. \] Step 2: First derivative of \( y \)
Differentiating both sides with respect to \( x \): \[ y' = \frac{d}{dx} \left( \frac{1}{1 - x} \right). \] Using the derivative of a rational function: \[ y' = \frac{1}{(1-x)^2}. \] Step 3: Second derivative of \( y \)
Differentiating again: \[ y'' = \frac{d}{dx} \left( \frac{1}{(1-x)^2} \right). \] Using the chain rule: \[ y'' = \frac{2}{(1-x)^3}. \] Step 4: Expressing \( y'' \) in terms of \( y \)
Since we have: \[ y = \frac{1}{1 - x}, \] \[ y' = \frac{1}{(1 - x)^2}, \] \[ y'' = \frac{2}{(1 - x)^3}. \] Rewriting \( y'' \) in terms of \( y \) and \( y' \): \[ y'' = 2 y y'. \] Step 5: Conclusion
Thus, the final answer is: \[ \boxed{2yy'}. \]