Key Concepts and Formulas

• Chain Rule for Differentiation: If $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.
• Differential Equation $f'(x) = f(x)$: The general solution to this differential equation is $f(x) = Ce^x$, where $C$ is a constant.
• Using Initial Conditions: An initial condition (like $f(1) = 2$) is used to determine the specific value of the constant in the general solution.

Step-by-Step Solution

Step 1: Solve the differential equation $f'(x) = f(x)$ We are given that $f'(x) = f(x)$ for all $x \in \mathbb{R}$. This is a first-order linear homogeneous differential equation. We can rewrite it as: $\frac{f'(x)}{f(x)} = 1$ Integrating both sides with respect to $x$: $\int \frac{f'(x)}{f(x)} dx = \int 1 dx$ Let $u = f(x)$, so $du = f'(x) dx$. The left side becomes $\int \frac{1}{u} du = \ln|u| = \ln|f(x)|$. The right side is $x + C_1$, where $C_1$ is the constant of integration. So, $\ln|f(x)| = x + C_1$. Exponentiating both sides: $|f(x)| = e^{x+C_1} = e^{C_1} e^x$ $f(x) = \pm e^{C_1} e^x$ Let $C = \pm e^{C_1}$. Since $f(x)$ is differentiable and $f'(x) = f(x)$, $f(x)$ must be continuous. If $f(x)$ were ever zero, then $f'(x)$ would also be zero at that point, implying a constant function of 0. However, we have an initial condition that prevents this. Therefore, $f(x)$ is never zero. Thus, we can write the general solution as $f(x) = Ce^x$.

Step 2: Use the initial condition to find the specific function $f(x)$ We are given that $f(1) = 2$. Substituting this into our general solution $f(x) = Ce^x$: $f(1) = Ce^1 = 2$ $Ce = 2$ $C = \frac{2}{e}$ Therefore, the specific function is: $f(x) = \frac{2}{e} e^x = 2e^{x-1}$

Step 3: Find the derivative of $f(x)$ Now that we have $f(x)$, we can find its derivative, $f'(x)$. $f(x) = 2e^{x-1}$ $f'(x) = \frac{d}{dx}(2e^{x-1}) = 2 \cdot e^{x-1} \cdot \frac{d}{dx}(x-1) = 2e^{x-1} \cdot 1 = 2e^{x-1}$ As a check, notice that $f'(x) = 2e^{x-1} = f(x)$, which is consistent with the given information.

Step 4: Apply the Chain Rule to find $h'(x)$ We are given $h(x) = f(f(x))$. To find $h'(x)$, we use the chain rule: $h'(x) = f'(f(x)) \cdot f'(x)$

Step 5: Evaluate $h'(1)$ We need to find $h'(1)$. Substitute $x=1$ into the expression for $h'(x)$: $h'(1) = f'(f(1)) \cdot f'(1)$ We know $f(1) = 2$ from the problem statement. We also need to find $f'(1)$. Using $f'(x) = 2e^{x-1}$: $f'(1) = 2e^{1-1} = 2e^0 = 2 \cdot 1 = 2$ Now, we need to evaluate $f'(f(1))$, which is $f'(2)$ since $f(1)=2$. Using $f'(x) = 2e^{x-1}$: $f'(2) = 2e^{2-1} = 2e^1 = 2e$ Substitute these values back into the expression for $h'(1)$: $h'(1) = f'(2) \cdot f'(1) = (2e) \cdot (2)$ $h'(1) = 4e$

Common Mistakes & Tips

• Incorrectly solving the differential equation: Ensure you correctly integrate $\frac{f'(x)}{f(x)}$ and use the initial condition to find the constant.
• Forgetting the chain rule: When differentiating a composite function like $f(f(x))$, remember to multiply by the derivative of the inner function.
• Calculation errors: Double-check the evaluation of $f(1)$, $f'(1)$, and $f'(f(1))$ to avoid simple arithmetic mistakes.

Summary

The problem requires us to find the derivative of a composite function $h(x) = f(f(x))$ at a specific point $x=1$. We are given a differential equation $f'(x) = f(x)$ and an initial condition $f(1) = 2$. First, we solved the differential equation to find the explicit form of $f(x) = 2e^{x-1}$. Then, we found its derivative, $f'(x) = 2e^{x-1}$. Using the chain rule, we expressed $h'(x)$ as $f'(f(x)) \cdot f'(x)$. Finally, we evaluated $h'(1)$ by substituting the known values of $f(1)$ and $f'(1)$, and calculating $f'(f(1)) = f'(2)$, which led to the result $4e$.

The final answer is \boxed{4e}, which corresponds to option (A).