Question

If $\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \cdot \mathrm{f}(\mathrm{z})\( for all \)\mathrm{x}, \mathrm{y}, \mathrm{z}$ and \(f(2)=4, f^{\prime}(0)=3\), then \(f^{\prime}(2)\) equals (A) 12 (B) 9 (C) 16 (D) 6

Short answer

Answer: The derivative of the function at the point \(x = 2\) is \(f'(2) = 6\).

Step-by-step solution

Differentiate the given formula

We will use the chain rule to differentiate both sides of the equation \(f(x+y+z) = f(x) \cdot f(y) \cdot f(z)\) with respect to \(x\). For this, we let \(u = x+y+z\) and rewrite the function as \(f(u) = f(x) \cdot f(y) \cdot f(z)\). Now, differentiating both sides with respect to \(x\), we get: \(\frac{df(u)}{dx} = \frac{df(x)}{dx} \cdot f(y) \cdot f(z)\) Now we need to apply the chain rule to \(\frac{df(u)}{dx}\): \(\frac{df(u)}{dx} = \frac{df(u)}{du} \cdot \frac{du}{dx} = f'(u)\). So, we have: \(f'(x+y+z) = f'(x) \cdot f(y) \cdot f(z)\)

Apply the given conditions

We know that \(f(2) = 4\), so we can find \(f(0)\) by plugging in \(x = z = 0\) and \(y = 2\) into the equation \(f(x+y+z) = f(x) \cdot f(y) \cdot f(z)\): \(f(2) = f(0) \cdot f(2) \cdot f(0)\) \(4 = f(0) \cdot 4 \cdot f(0)\) Solving this equation, we get \(f(0)=\frac{1}{2}\). Now we need to find \(f'(2)\) by plugging in \(x = z = 0\) and \(y = 2\) into the equation \(f'(x+y+z) = f'(x) \cdot f(y) \cdot f(z)\): \(f'(2) = f'(0) \cdot f(2) \cdot f(0)\) Given that \(f'(0)=3\) and \(f(0)=\frac{1}{2}\), we plug in these values into the equation: \(f'(2) = 3 \cdot 4 \cdot \frac{1}{2}\)

Solve for \(f'(2)\)

Now that we've plugged in the given values, we just need to compute \(f'(2)\): \(f'(2) = 3 \cdot 4 \cdot \frac{1}{2} = 6\) So, the derivative at \(x=2\) is \(f'(2)=6\), which corresponds to option (D).