Question

How do you calculate the derivative of `(cos(t^2)+t) dt` from `t=-5` to `t=sinx`?

Answer 1

`= cos x( cos (sin x)^2 + sin x)`

Explanation:

i think you mean this

`d/dx int_{-5}^{sin x} \ cos t^2 + t\ dt`

We can use the Leibnitz Rule for differentiating under the integral sign

which simplifies considerably here to

`d/dx int_{-5}^{sin x} \ cos t^2 + t\ dt`

`= cos (sin x)^2 (sin x)' + sin x (sin x)'`

`= cos (sin x)^2 cos x + sin x cos x`

`= cos x( cos (sin x)^2 + sin x)`

if that seems overwhelming, we can look at the Fundamental Theorem of Calculus, specifically

if `F(x) = int_a^x f(t) \ dt`, then `F'(x) = f(x)`

this can be extended using the chain rule to

if `F(x) = int_a^{v(x)} f(t) \ dt`, then `F'(x) = f(v(x)) * v'(x)`

that's all we did here