Let f be a differentiable function such that f(1) = 2 and f'(x) = f(x) for all x ∈ R. If h(x) = f(f(

Q: Wait, why can't we just use energy conservation or some other shortcut?

Energy conservation is a physical principle, not a mathematical one! In calculus, we rely on the definitions of derivatives and composite functions. The Chain Rule is the only 'shortcut' you need here, as it directly connects the derivative of the composite function to the derivatives of its components.

Q: Why is $f'(2) = f(2)$? I thought the derivative was only equal to the function at $x=1$.

Great question! The problem states that $f'(x) = f(x)$ for *all* $x \in R$. This means the property holds true for any value of $x$, including $x=2$. It is a fundamental identity of this specific function, not just a condition at a single point.

Q: How do I know $f(x) = Ae^x$ is the only solution?

For the differential equation $\frac{df}{dx} = f$, separating variables gives $\int \frac{1}{f} df = \int dx$, which leads to $\ln|f| = x + C$. Exponentiating both sides yields $f(x) = e^{x+C} = Ae^x$. The initial condition $f(1) = 2$ allows us to solve for the unique constant $A$, ensuring there is only one specific function that satisfies both the differential equation and the initial condition.

Question

Let $f$ be a differentiable function such that $f(1) = 2$ and $f'(x) = f(x)$ for all $x \in R$. If $h(x) = f(f(x))$, then $h'(1)$ is equal to :

Options:

(A) $4e$

(B) $4e^2$

(C) $2e$

(D) $2e^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe! Today, we are going to unravel a problem that feels like a riddle wrapped in an enigma.

We are looking at a function $f(x)$ that possesses a rare and beautiful property: its derivative is identical to itself, $f'(x) = f(x)$. This is the mathematical embodiment of a process that feeds on its own magnitude.

We are also given an initial anchor, $f(1) = 2$, which pins our curve to a specific coordinate in the plane. Our mission is to find the derivative of a composite function, $h(x) = f(f(x))$, at the point $x=1$.

### The Divorce of Variables

To understand the soul of our function $f(x)$, we must first solve the differential equation $f'(x) = f(x)$. We can rewrite this as:

$$ \frac{df}{dx} = f $$

To solve this, we perform a separation of variables. By dividing both sides by $f$ and multiplying by $dx$, we obtain:

$$ \frac{1}{f} df = 1 dx $$

Now, we integrate both sides. The integral of $\frac{1}{f}$ is $\ln|f|$, and the integral of $1$ is $x$. Thus, we have $\ln|f(x)| = x + C$.

Exponentiating both sides, we find $f(x) = e^{x+C}$, which simplifies to $f(x) = Ae^x$, where $A$ is a constant. This is the general form of our function.

### The Russian Doll of Functions

Now, we turn our attention to the composite function $h(x) = f(f(x))$. Think of this as a Russian nesting doll: $f$ is inside $f$.

To find the derivative $h'(x)$, we must use the **Chain Rule**. It tells us that:

$$ h'(x) = f'(f(x)) \cdot f'(x) $$

We need to evaluate this at $x=1$, so we write:

$$ h'(1) = f'(f(1)) \cdot f'(1) $$

### The Final Calculation

We know $f(1) = 2$. Substituting this into our expression, we get $h'(1) = f'(2) \cdot f'(1)$.

Since $f'(x) = f(x)$, we know $f'(1) = f(1) = 2$. Furthermore, $f'(2) = f(2)$.

Using our general form $f(x) = Ae^x$ and the condition $f(1) = 2$, we find:

$$ 2 = Ae^1 \Rightarrow A = \frac{2}{e} $$

Thus, the specific function is $f(x) = \frac{2}{e} \cdot e^x = 2e^{x-1}$.

Now, we calculate $f(2)$:

$$ f(2) = 2e^{2-1} = 2e $$

Substituting everything back into our expression for $h'(1)$, we get:

$$ h'(1) = (2e) \cdot (2) = 4e $$

The elegance of this result, where the constants and variables dance together to form a clean, final answer, is the true reward of our labor. The final result is **$4e$**.

Animated Solution for Mathematics - Differential Equations: Let $f$ be a differentiable function such that $f (1) = 2$ and $f^{'} (x) = f (x)$ for all $x \in R$ . If $h (x) = f (f (x))$ , then $h^{'} (1)$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

Setting up the Differential Equation

Integrating to find $f (x)$

Applying the Initial Condition

Finalizing the Function $f (x)$

The Chain Rule for $h (x)$

Substituting $x = 1$

Evaluating $f (1)$ and $f^{'} (1)$

Calculating $f^{'} (2)$

Final Computation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Divorce of Variables

The Russian Doll of Functions

The Final Calculation

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Visualized Solution

Visualizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Setting up the Differential Equation

Integrating to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Applying the Initial Condition

Finalizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Chain Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(1)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(2)