If ` f'(1) = 2 and v = f (log_(e)x) ` , then `( dv)/(dx) ` is for x = e

Options

A. `(2)/( e)`
B. 0
C. e
D. 1

Answer

To solve the problem, we need to find \(\frac{dv}{dx}\) given that \(v = f(\log_e x)\) and \(f'(1) = 2\). We will use the chain rule for differentiation.

### Step-by-step Solution:

1. **Identify the function**: We have \(v = f(\log_e x)\). Here, \(f\) is a function of \(\log_e x\).

2. **Apply the Chain Rule**: To find \(\frac{dv}{dx}\), we will apply the chain rule. According to the chain rule:
\[ \frac{dv}{dx} = f'(\log_e x) \cdot \frac{d}{dx}(\log_e x) \]

3. **Differentiate \(\log_e x\)**: The derivative of \(\log_e x\) with respect to \(x\) is:
\[ \frac{d}{dx}(\log_e x) = \frac{1}{x} \]

4. **Substitute into the chain rule**: Now we can substitute this back into our expression for \(\frac{dv}{dx}\):
\[ \frac{dv}{dx} = f'(\log_e x) \cdot \frac{1}{x} \]

5. **Evaluate at \(x = e\)**: We need to find \(\frac{dv}{dx}\) when \(x = e\):
- First, calculate \(\log_e e\):
\[ \log_e e = 1 \]
- Now substitute \(x = e\) into the derivative:
\[ \frac{dv}{dx} = f'(\log_e e) \cdot \frac{1}{e} = f'(1) \cdot \frac{1}{e} \]

6. **Use the given value of \(f'(1)\)**: We know from the problem that \(f'(1) = 2\):
\[ \frac{dv}{dx} = 2 \cdot \frac{1}{e} = \frac{2}{e} \]

### Final Answer:
\[ \frac{dv}{dx} \text{ at } x = e \text{ is } \frac{2}{e} \]

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