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 given problem, we need to determine the value of \(\frac{h'(x)}{h(x)}\) where \(h(x) = (f \circ f)(x) = f(f(x))\) and \(f(x) = e^x\).

First, let's define the function \(h(x)\):

\[ h(x) = f(f(x)) = f(e^x) = e^{e^x} \]

Now, we need to find the derivative \(h'(x)\):

\[ h'(x) = \frac{d}{dx}[e^{e^x}] \]

Using the chain rule, the derivative of \(e^{e^x}\) is:

1. Inner function derivative: \(e^x\) is \(\frac{d}{dx}[e^x] = e^x\)

2. Outer function derivative: \(e^u\) where \(u = e^x\) is \(e^u \times \frac{du}{dx}\)

Thus, the derivative \(h'(x)\) is:

\[ h'(x) = e^{e^x} \cdot e^x = e^{e^x + x} \]

Next, substituting \(h(x) = e^{e^x}\) and \(h'(x) = e^{e^x + x}\) into \(\frac{h'(x)}{h(x)}\), we have:

\[ \frac{h'(x)}{h(x)} = \frac{e^{e^x + x}}{e^{e^x}} = e^x \]

From here, we realize that our goal is to match \(\frac{h'(x)}{h(x)} = \log h(x)\):

\[ \log h(x) = \log(e^{e^x}) = e^x \]

Thus, \(\frac{h'(x)}{h(x)} = \log h(x)\). Therefore, the correct answer is: \( \log h(x) \).

Hence, the correct choice is:

\(log h(x)\)