Question

I.et \(\mathrm{f}(\mathrm{x})\) be a function such that \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})\) and \(f(x)=\sin x g(x)\) for all \(x, y \in R\), If \(g(x)\) is a continuous function such that \(\mathrm{g}(0)=\mathrm{K}\), then \(f^{\prime}(\mathrm{x})\) is equal to (A) \(\mathrm{K}\) (B) \(\mathrm{Kx}\) (C) \(\mathrm{Kg}(\mathrm{x})\) (D) none

Short answer

Based on the given conditions and the functional equation, find the expression for the derivative of the function \(f'(x)\). Answer: The derivative of the function, \(f'(x)\), is given by the expression \(\cos x \cdot (x + K) + \sin x\). This does not match any of the given answer choices (A) K, (B) Kx, or (C) Kg(x), so the answer is (D) none.

Step-by-step solution

Solve the functional equation

Given the functional equation: \(f(x+y) = f(x) + f(y)\). We can differentiate both sides of the equation with respect to \(x\) to find the derivative of \(f(x+y)\) with respect to \(x\). To do this, we use the Chain Rule: \(\frac{d}{dx}(f(x+y)) = \frac{d}{dx}(f(x) + f(y))\) Since \(y\) is independent of \(x\), we can treat \(f(y)\) as a constant when differentiating with respect to \(x\). Therefore, we have: \(f'(x+y) = f'(x)\) Now, we can rewrite the expression for \(f(x)\) using the information given: \(f(x) = \sin x \cdot g(x)\)

Differentiate \(f(x)\) with respect to \(x\)

We can now differentiate \(f(x)\) with respect to \(x\) using the Product Rule: \(f'(x) = \frac{d}{dx}(\sin x \cdot g(x)) = (\sin x)' \cdot g(x) + \sin x \cdot g'(x)\) We know the derivative of \(\sin x\) is \(\cos x\), so the expression becomes: \(f'(x) = \cos x \cdot g(x) + \sin x \cdot g'(x)\) Using the results from Step 1, we now substitute for \(f'(x+y)\): \(f'(x+y) = \cos (x+y) \cdot g(x+y) + \sin (x+y) \cdot g'(x+y)\)

Use the relationship between \(f'(x+y)\) and \(f'(x)\)

From Step 1, we know that \(f'(x+y) = f'(x)\), so we can set the expressions for \(f'(x+y)\) and \(f'(x)\) equal: \(\cos (x+y) \cdot g(x+y) + \sin (x+y) \cdot g'(x+y) = \cos x \cdot g(x) + \sin x \cdot g'(x)\) Since this equality must hold for all \(x\) and \(y\), we can set the \(x\) and \(y\) terms equal after substituting \(x = 0\): \(0 \cdot g(y) + K \cdot g'(y) = K\) Since this equation must hold for all \(y\), we have that \(g'(y) = 1\). If we integrate \(g'(y)\), we get \(g(y) = y + c\) for some constant \(c\). Use the given condition \(g(0) = K\) to determine the value of \(c\). Therefore, \(c=K\) and we get: \(g(y) = y + K\)

Compute the derivative \(f'(x)\)

To find the derivative \(f'(x)\), we can substitute the expression \(g(y) = y + K\) into our previously derived expression for \(f'(x)\): \(f'(x) = \cos x \cdot g(x) + \sin x \cdot g'(x) = \cos x \cdot (x + K) + \sin x \cdot 1\) Since none of the given answer choices (A) K, (B) Kx, or (C) Kg(x) match the expression we found for \(f'(x)\), the correct answer must be (D) none.