how to find the derivative of the function. y=integral cos(u^4) from 4cos(x) to 4x

Answer 1

integral cos u^4) du = (1/4) sin u^4
at u = 4x integral = (1/4) sin (4x)^4
at u = 4 cos x integral = (1/4) sin (4 cos x)^4
so definite integral = (1/4)[ sin (4x)^4 - sin (4 cos x)^4 ]
d/dx of that
= (1/4)[ 16 sin (4x)^3 - 4 sin(4cosx)^3 (-4sin x) ]
etc

Answer 2

The derivative F'(x) of the integral
F(x) = ∫[a(x),b(x)] f(u) du = F(b(x))*b'(x) - f(a(x))*a'(x)
with
a(x)=cos(4x)
b(x)=4x
f(u)=cos(u^4)
F'(x)= cos((4cosx)^4)*(4cosx)^3*(-4sinx) - (cos(4x)^4)*4(4x)^3*4