Question

If \( y = e^{(x+e)^{(x+e)^{(x+\cdots)}}} \), what is the value of \( \frac{d}{dx}(y) \)?

• A. \( \frac{d}{dx}(y) = \frac{y}{1 - y} \)
• B. \( \frac{d}{dx}(y) = \frac{y}{1 + y} \)
• C. \( \frac{d}{dx}(y) = \frac{1 - y}{1 + y} \)
• D. \( \frac{d}{dx}(y) = \frac{1 + y}{1 - y} \)

Hint:

For recursive functions, use the chain rule and express the recursive part as a new variable to simplify differentiation.

Answer 1

The Correct Option is B

Step 1: Differentiate the function.
We are given a recursive exponential function. To differentiate it, we use the chain rule. \[ y = e^{(x+e)^{(x+e)^{(x+\cdots)}}} \] Let \( z = (x+e)^{(x+e)^{(x+\cdots)}} \), so \( y = e^z \). Step 2: Apply the chain rule.
Using the chain rule, we differentiate \( y = e^z \) with respect to \( x \), and we arrive at the result \( \frac{d}{dx}(y) = \frac{y}{1 + y} \). Final Answer: \[ \boxed{\frac{y}{1 + y}} \]