Answer: Choice D
[tex]\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\[/tex]
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Explanation:
Let g(t) be the antiderivative of [tex]g'(t) = \sqrt{1+t^3}[/tex]. We don't need to find out what g(t) is exactly.
Recall by the fundamental theorem of calculus, we can say the following:
[tex]\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)[/tex]
This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).
So,
[tex]\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\[/tex]
From here, we apply the derivative with respect to x to both sides. Note that the [tex]g(\pi)[/tex] portion is a constant, so [tex]g'(\pi) = 0[/tex]
[tex]\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\[/tex]
[tex]\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\[/tex]
Answer is choice D
[tex]F(x)=\displaystyle\int_\pi^{x^2}\sqrt{1+t^3}dt~\hfill u = x^2\implies \cfrac{du}{dx}=2x\\\\\\\cfrac{dF}{du}\cdot \cfrac{du}{dx}\implies \cfrac{dF}{dx}~\hfill \cfrac{d}{dx}\left[ \displaystyle\int_\pi^{x^2}\sqrt{1+t^3}dt\right]\implies \cfrac{d}{du}\left[ \displaystyle\int_\pi^{u}\sqrt{1+t^3}dt\right]\cfrac{du}{dx}\\\\\\\sqrt{1+u^3}\cdot \cfrac{du}{dx}\implies \sqrt{1+u^3}2x\implies 2x\sqrt{1+(x^2)^3}\implies \boxed{2x\sqrt{1+x^6}}[/tex]
