Pls help me Im struggling to do this.

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

7 0

Answer:

The first answer is that the exponents decrease by one in a patern

Step-by-step explanation:

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Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$