Find the derivative of the function

y=S 10(t)^1/2 sin(t)dt
y'=?
S=integration symbol
b=x^3
a=(x)^1/2

at b = x^3

the change of the integral with x = y'(x) [ it is the value of the function at that limit ]
=10 x^3/2 sin( x^3 )

as with the other problem,

y = ∫[√x,x^3] 10√t sin(t)dt
y' = (10√(x^3) sin(x^3))(3x^2) - (10√(√x) sin(√x))(1/(2√x))
= 30x7/2 sin(x^3) - 5/∜x sin(√x)

To find the derivative of the given function, you need to apply the Fundamental Theorem of Calculus.

Firstly, let's rewrite the integral in a more familiar notation. The notation used with "S" represents the integral symbol (∫), and "dt" indicates that we are integrating with respect to the variable t.

The function you provided can be written as:

y = ∫(from x^(1/2) to x^3) 10t^(1/2)sin(t) dt

To find the derivative, we will use the Second Fundamental Theorem of Calculus, which states that if F(x) is an antiderivative of f(x), then the derivative of the definite integral from a to x of f(t) dt is given by F(x) - F(a).

In this case, we can focus on finding the antiderivative of the function inside the integral, which is 10t^(1/2)sin(t).

To find the antiderivative, integrate each term separately. The antiderivative of 10t^(1/2) can be found using the power rule of integration, which gives us (20/3)t^(3/2). The antiderivative of sin(t) is -cos(t).

Therefore, the antiderivative of 10t^(1/2)sin(t) is (20/3)t^(3/2)(-cos(t)).

Now, we will substitute the limits of integration into the antiderivative we just found:

F(x) - F(a) = [(20/3)t^(3/2)(-cos(t))] evaluated from x^(1/2) to x^3

To compute the derivative, we subtract the result with the lower limit (x^(1/2)) from the result with the upper limit (x^3):

y' = [(20/3)(x^3)^(3/2)(-cos(x^3))] - [(20/3)(x^(1/2))^(3/2)(-cos(x^(1/2)))]

Simplifying this expression gives us the derivative of the function y with respect to x.