Question

Use the Product Rule to differentiate the function. $$ g(x)=\sqrt{x} \sin x $$

Short answer

The derivative of the function \(g(x) = \sqrt{x}\sin{x}\) is \(g'(x) = \frac{\sin{x}}{2\sqrt{x}} + \sqrt{x} \cdot \cos{x}\).

Step-by-step solution

Apply the Product Rule

We start by applying the Product Rule, which states that the derivative of \(f(x) \cdot g(x)\) is \(f'(x) \cdot g(x) + f(x) \cdot g'(x)\). In this case, \(f(x) = \sqrt{x} = x^{1/2}\) and \(g(x) = \sin{x}\). So we need to first compute \(f'(x)\) and \(g'(x)\), and then use them in the Product Rule formula.

Compute the Derivatives of \(f(x)\) and \(g(x)\)

First, the derivative of \(f(x) = x^{1/2}\) is obtained using the power rule. So, \(f'(x) = \frac{1}{2}x^{-1/2}\). And the derivative of \(g(x) = \sin{x}\) is simply \(g'(x) = \cos{x}\). We can now substitute these derivatives back into the Product Rule formula.

Substitute Back into the Product Rule Formula

Substituting these values into our formula, we get: \(g'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) = \frac{1}{2}x^{-1/2} \cdot \sin{x} + x^{1/2} \cdot \cos{x}\).

Simplify the Result

Our result can be simplified by writing \(x^{-1/2}\) as \(1/\sqrt{x}\), yielding the final answer \(g'(x) = \frac{\sin{x}}{2\sqrt{x}} + \sqrt{x} \cdot \cos{x}\).