**Differentiating √xsinx sin√x**

To differentiate the expression √xsinx sin√x, we can use the product rule of differentiation. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:

(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)

In this case, we can identify u(x) = √xsinx and v(x) = sin√x. Let's differentiate each of these functions separately and then apply the product rule to find the derivative of the expression.

**Differentiating √xsinx**

To differentiate √xsinx, we need to apply the chain rule since we have functions nested inside each other. The chain rule states that if we have a composite function f(g(x)), then the derivative of f(g(x)) is given by:

(d/dx)(f(g(x))) = f'(g(x)) * g'(x)

In this case, let's consider f(u) = √u and g(x) = sinx. The derivative of f(u) = √u is given by f'(u) = (1/2√u). The derivative of g(x) = sinx is g'(x) = cosx.

Now, we can apply the chain rule:

(d/dx)(√xsinx) = f'(g(x)) * g'(x)
= (1/2√u) * cosx
= (1/2√(sinx)) * cosx

So, the derivative of √xsinx is (1/2√(sinx)) * cosx.

**Differentiating sin√x**

To differentiate sin√x, we can use the chain rule again. Let's consider f(u) = sinu and g(x) = √x. The derivative of f(u) = sinu is f'(u) = cosu. The derivative of g(x) = √x is g'(x) = (1/2√x).

Now, we can apply the chain rule:

(d/dx)(sin√x) = f'(g(x)) * g'(x)
= cos(√x) * (1/2√x)
= (1/2√x) * cos(√x)

So, the derivative of sin√x is (1/2√x) * cos(√x).

**Applying the Product Rule**

Now that we have found the derivatives of √xsinx and sin√x, we can apply the product rule to find the derivative of √xsinx sin√x.

(d/dx)(√xsinx sin√x) = (√xsinx)' * sin√x + √xsinx * (sin√x)'
= [(1/2√(sinx)) * cosx] * sin√x + √xsinx * [(1/2√x) * cos(√x)]

Simplifying this expression further is not possible without additional information or constraints on x. So, the final derivative of √xsinx sin√x is [(1/2√(sin