Question

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

Answer 1

Let `F(x)=\int_{2}^{x}\sqrt{5t+\sqrt{t}}\ dt` and `g(x)=\tan(x)`. The Fundamental Theorem of Calculus implies that `F'(x)=\sqrt{5x+\sqrt{x}}`. The derivative of the tangent function is `g'(x)=\sec^{2}(x)`.

Since `\int_{2}^{\tan(x)}\sqrt{5t+\sqrt{t}}\ dt=F(g(x))`,
the Chain Rule can now be used to say that

`d/(dx}(\int_{2}^{\tan(x)}\sqrt{5t+\sqrt{t}}\ dt)=F'(g(x))*g'(x)`

`=\sqrt{5\tan(x)+\sqrt{\tan(x)}}*\sec^{2}(x)`.