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: Function Differentiation.
Given the recursive exponential function, we apply the chain rule for differentiation.\[y = e^{(x+e)^{(x+e)^{(x+\cdots)}}}\]Define \( z = (x+e)^{(x+e)^{(x+\cdots)}} \), thus \( y = e^z \).Step 2: Chain Rule Application.
Applying the chain rule to differentiate \( y = e^z \) with respect to \( x \), we obtain \( \frac{d}{dx}(y) = \frac{y}{1 + y} \). Final Answer: \[ \boxed{\frac{y}{1 + y}} \]