find the derivative of the function

y= integral from square root of x, to x cubed of (square root t * sin(t))dt

To find the derivative of the function, y = ∫√(x) to x^3 (√t * sin(t))dt, we can use the Fundamental Theorem of Calculus. The theorem states that if a function, f(x), is continuous on the interval [a, x], then the derivative of the integral with respect to x is equal to the function evaluated at the upper limit of integration, x.

Let's denote the integral as F(x) = ∫√(x) to x^3 (√t * sin(t))dt. We need to substitute the upper limit of integration, x^3, into the function (√t * sin(t)) and differentiate with respect to x.

To begin, we can break down the integral into two separate integrals. First, we integrate from √(x) to t and then from t to x^3.

F(x) = ∫√(x) to x^3 (√t * sin(t))dt
= ∫√(x) to t (√t * sin(t))dt + ∫t to x^3 (√t * sin(t))dt

Now, using the Fundamental Theorem of Calculus, we can differentiate the integral with respect to x.

d/dx [F(x)] = d/dx ∫√(x) to t (√t * sin(t))dt + d/dx ∫t to x^3 (√t * sin(t))dt

The first term requires using the Chain Rule and the lower limit of integration, which is a function of x.

d/dx ∫√(x) to t (√t * sin(t))dt = (√t * sin(t))(dt/dx)
= (√t * sin(t))(d/dt t)

Since t is a variable of integration (a dummy variable), differentiating t with respect to t gives us 1.

d/dx ∫√(x) to t (√t * sin(t))dt = √t * sin(t)

Now, applying the Fundamental Theorem of Calculus to the second part:

d/dx ∫t to x^3 (√t * sin(t))dt = (√t * sin(t))(dt/dx)
= (√t * sin(t))(d/dt t^3)

Again, differentiating t^3 with respect to t gives us 3t^2.

d/dx ∫t to x^3 (√t * sin(t))dt = √t * sin(t) * 3t^2

Combining both terms, we have:

d/dx [F(x)] = √t * sin(t) + √t * sin(t) * 3t^2

Therefore, the derivative of the function y = ∫√(x) to x^3 (√t * sin(t))dt with respect to x is given by:

dy/dx = √t * sin(t) + √t * sin(t) * 3t^2

Note that we cannot simplify the expression further since t is bound within the integral.