Question

If \(f(x)\) is a differentiable function and \[ y=e^{f(x)+e^{f(x)+e^{f(x)+\cdots \infty}}}, \] then \[ \frac{dy}{dx}= \ ? \]

• A. \(\dfrac{yf'(x)}{1+y}\)
• B. \(\dfrac{(1+y)f'(x)}{y}\)
• C. \(\dfrac{(1-y)f'(x)}{y}\)
• D. \(\dfrac{yf'(x)}{1-y}\)

Hint:

Whenever an expression repeats infinitely, assign it a variable and use self-similarity to create an equation.

Answer 1

The Correct Option is D

Concept: For an infinite nested exponential expression, we first represent the entire expression by a variable and then form an equation involving that variable. Implicit differentiation is then used.

Step 1: Express the infinite exponential recursively.
Given \[ y=e^{f(x)+e^{f(x)+e^{f(x)+\cdots}}}. \] The infinite tail is again equal to \(y\). Therefore, \[ y=e^{f(x)+y}. \]

Step 2: Take logarithm on both sides.
\[ \ln y=f(x)+y. \]

Step 3: Differentiate implicitly.
Differentiating both sides with respect to \(x\), \[ \frac1y\frac{dy}{dx} = f'(x)+\frac{dy}{dx}. \] Rearranging, \[ \frac{dy}{dx} \left( \frac1y-1 \right) = f'(x). \]

Step 4: Solve for \(\frac{dy}{dx}\).
\[ \frac{dy}{dx} = \frac{f'(x)} {\frac{1-y}{y}}. \] Hence \[ \frac{dy}{dx} = \frac{yf'(x)}{1-y}. \] Therefore, \[ \boxed{\frac{yf'(x)}{1-y}}. \]