Question

If f(x) = e^x, h(x) = (fof) (x), then \(\frac{h'(x)}{h'(x)}\) = 

• A. h(x)
• B. \(\frac{1}{h(x)}\)
• C. \(log h(x)\)
• D. \(-log h(x)\)

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate $\frac{h'(x)}{h(x)}$ for the function $h(x) = (f \circ f)(x)$, where $f(x) = e^x$

1. Define the Function $h(x)$:
Given $f(x) = e^x$, we have $h(x) = (f \circ f)(x) = f(f(x)) = f(e^x) = e^{e^x}$.

2. Compute the Derivative $h'(x)$:
Differentiate $h(x) = e^{e^x}$ using the chain rule:
$h'(x) = e^{e^x} \cdot \frac{d}{dx}(e^x) = e^{e^x} \cdot e^x = e^x \cdot e^{e^x} = e^x \cdot h(x)$.

3. Evaluate $\frac{h'(x)}{h(x)}$:
Using the result from step 2:
$\frac{h'(x)}{h(x)} = \frac{e^x \cdot h(x)}{h(x)} = e^x$.

4. Analyze the Options:
Option 1: $h(x) = e^{e^x}$, which is the definition of $h(x)$, not related to $\frac{h'(x)}{h(x)}$.
Option 2: $\frac{1}{h(x)} = \frac{1}{e^{e^x}} = e^{-e^x}$, which does not equal $e^x$.
Option 3: $\log h(x) = \log(e^{e^x}) = e^x$, which matches $\frac{h'(x)}{h(x)} = e^x$.
Thus, $\frac{h'(x)}{h(x)} = \log h(x)$.

Final Answer:
The correct option is $\log h(x)$.