let g(x)=integral x to (1/2) square root (t^3+1)dt .find g'(x)

Answer 1

take a look at wikipedia under differentiation inder the integral sign to see that

g'(x) = -sqrt(x^3+1)

On the web page, f(x,t) becomes just f(t), so partial df/dx = 0; b(x) is a constant, so b'=0.