Question

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

Answer 1

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 )

Answer 2

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))
= 30x^{^7/2} sin(x^3) - 5/∜x sin(√x)