Fundamental Theorem of Calculus Proof

Question

Find $f'$ where is $f$ is defined on $[0, 1]$ as indicated: $$f(x) = \int_x^{\sqrt{x}} \frac 1{1+t^3}dt$$ I know that the fundamental theorem is going to be used in this proof, but I'm not really sure where to begin. So any help would be greatly appreciated. Thank you!

Answer 1

Let $g'(x)=\frac{1}{1+x^3}$, then by the FTC, $$f(x) = \int_x^{\sqrt{x}} g'(t)\,dt=g(\sqrt x)-g(x)$$ then $$f'(x)=\frac{g'(\sqrt x)}{2\sqrt x}-g'(x)=\frac{1}{2(\sqrt x+x^2)}-\frac{1}{1+x^3}$$

Answer 2

We have: $f(x) = \displaystyle \int_{0}^{\sqrt{x}} \dfrac{1}{1+t^3} dt - \displaystyle \int_{0}^x \dfrac{1}{1+t^3} dt$. Apply the fundamental theorem of calculus and the chainrule we arrive at..:

$f'(x) = \dfrac{1}{2\sqrt{x}\cdot (1+x\sqrt{x})} - \dfrac{1}{1+x^3} = \dfrac{1}{2\sqrt{x} + 2x^2} - \dfrac{1}{1+x^3}$.