Let `F (x) = e ^(x) , G (x) =e ^(-x) and H (x) = G (F(x)),` where x is a real variable. Then `(dH)/(dx) at x =0` is

Options

A. 1
B. `-1`
C. `-1/e`
D. `-e`

Answer

To solve the problem, we need to find the derivative of the function \( H(x) = G(F(x)) \) at \( x = 0 \). Let's break this down step by step.

### Step 1: Define the functions
We have:
- \( F(x) = e^x \)
- \( G(x) = e^{-x} \)

### Step 2: Substitute \( F(x) \) into \( G(x) \)
Now, we can express \( H(x) \):
\[ H(x) = G(F(x)) = G(e^x) = e^{-e^x} \]

### Step 3: Differentiate \( H(x) \)
To find \( \frac{dH}{dx} \), we will use the chain rule. The chain rule states that if you have a composite function \( H(x) = G(F(x)) \), then:
\[ \frac{dH}{dx} = G'(F(x)) \cdot F'(x) \]

#### Step 3.1: Find \( F'(x) \)
The derivative of \( F(x) = e^x \) is:
\[ F'(x) = e^x \]

#### Step 3.2: Find \( G'(x) \)
The derivative of \( G(x) = e^{-x} \) is:
\[ G'(x) = -e^{-x} \]

### Step 4: Substitute into the derivative formula
Now substituting into the chain rule formula:
\[ \frac{dH}{dx} = G'(F(x)) \cdot F'(x) = -e^{-F(x)} \cdot e^x \]
Substituting \( F(x) = e^x \):
\[ \frac{dH}{dx} = -e^{-e^x} \cdot e^x \]

### Step 5: Evaluate at \( x = 0 \)
Now we need to evaluate \( \frac{dH}{dx} \) at \( x = 0 \):
\[ \frac{dH}{dx} \bigg|_{x=0} = -e^{-e^0} \cdot e^0 = -e^{-1} \cdot 1 = -\frac{1}{e} \]

### Final Answer
Thus, the value of \( \frac{dH}{dx} \) at \( x = 0 \) is:
\[ \frac{dH}{dx} \bigg|_{x=0} = -\frac{1}{e} \]

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