Question

How do you use part 1 of the fundamental theorem of calculus to find the derivative of `g(x)= int cos(sqrt( t))dt` 6 to x?

Answer 1

The theorem says:

Given the function:

`y=int_(h(x))^g(x)f(t)dt`

then:

`y'=f(g(x))*g'(x)-f(h(x))*h'(x)`.

So:

`g'(x)=cossqrtx*1-cossqrt6*0=cossqrtx`.