let the function ln(f(x)) is defined where f(x) exists for x>2 and k is fixed positive real number, prove that if d/dx (x.f(x))

To prove that the derivative of the function \( x \cdot f(x) \) exists under the given conditions, we need to analyze the function \( f(x) \) and its properties. We know that \( \ln(f(x)) \) is defined for \( x > 2 \), which implies that \( f(x) \) must be positive and differentiable in that interval. Let's break down the proof step by step.

Understanding the Function

Given that \( \ln(f(x)) \) is defined, we can infer that \( f(x) > 0 \) for \( x > 2 \). This is crucial because the logarithm function is only defined for positive arguments. We also assume that \( f(x) \) is differentiable, which is a common requirement in calculus.

Applying the Product Rule

To find the derivative of the product \( x \cdot f(x) \), we can use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

• \( \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \)

In our case, let \( u(x) = x \) and \( v(x) = f(x) \). Therefore, we can differentiate \( x \cdot f(x) \) as follows:

Calculating the Derivative

Applying the product rule:

• \( \frac{d}{dx}(x \cdot f(x)) = \frac{d}{dx}(x) \cdot f(x) + x \cdot \frac{d}{dx}(f(x)) \)
• \( = 1 \cdot f(x) + x \cdot f'(x) \)
• \( = f(x) + x \cdot f'(x) \)

This shows that the derivative \( \frac{d}{dx}(x \cdot f(x)) \) exists as long as \( f(x) \) is differentiable, which we assumed earlier.

Conclusion on Existence

Since \( f(x) \) is positive and differentiable for \( x > 2 \), the expression \( f(x) + x \cdot f'(x) \) is well-defined and exists for all \( x > 2 \). Therefore, we have proven that the derivative \( \frac{d}{dx}(x \cdot f(x)) \) exists under the given conditions.

Final Thoughts

This proof highlights the importance of understanding the properties of logarithmic and differentiable functions. The product rule is a powerful tool in calculus that allows us to differentiate products of functions easily. If you have any further questions or need clarification on any part of this proof, feel free to ask!