Question

How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function of `y=int sqrt(2t+sqrt(t))dt` from 5 to `tanx`?

Answer 1

To find the derivative of the function of `y=int_5^tanx sqrt(2t+sqrt(t))dt`, we have to pair Part 1 of the Fundamental Theorem of Calculus with the Chain Rule.

FTC Part I tells us that for `g(x)= int_a^x f(t) dt`, the derivative with respect to `x` is `g'(x) = f(x)`

Now we combine this with the chain rule:

For the function `g(u)= int_a^u f(t) dt`, the derivative with respect to `x` is

`d/dx(g(u)) = g'(u) (du)/dx = f(u) (du)/dx`.

So here's the answer to this question:

For `y=int_5^tanx sqrt(2t+sqrt(t))dt`, the derivative with respect to `x` is

`dy/dx = sqrt(2tanx +sqrt(tanx)) * sec^2x`.