MHB Do's question at Yahoo Answers regarding differentiating a definite integral

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To find the derivative of the function defined by the integral from x^(1/2) to 1 of (s^2)/(1+3s^4) ds, the Fundamental Theorem of Calculus is applied. The function is expressed as F(x) = ∫(from √x to 1) (s^2)/(1+3s^4) ds. The derivative is calculated using the chain rule, leading to F'(x) = -√x/(2(1+3x^2)). This approach effectively utilizes the anti-derivative form of the theorem and the properties of derivatives. The final result provides a clear solution to the original question posed.
MarkFL
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Here is the question:

Find the derivative of the following function?

( ∫ )(from x^(1/2) to 1) ((s^2)/(1+3s^4))ds using the appropriate form of the Fundamental Theorem of Calculus. F'(x) =?

I have posted a link there to this topic so the OP can see my work.
 
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Hello do,

We are given to evaluate:

$$F(x)=\int_{\sqrt{x}}^1\frac{s^2}{1+3s^4}\,ds$$

Let:

$$f(s)=\frac{d}{ds}F(s)=\frac{s^2}{1+3s^4}$$

Using the anti-derivative form of the fundamental theorem of calculus, we may write:

$$f(x)=\frac{d}{dx}F(x)=\frac{d}{dx}\left(F(1)-F\left(\sqrt{x} \right) \right)$$

Using the fact that the derivative of a constant is zero on the first term, and applying the chain rule on the second term, we find:

$$f(x)=0-\frac{\left(\sqrt{x} \right)^2}{1+3\left(\sqrt{x} \right)^4}\frac{d}{dx}\left(\sqrt{x} \right)=-\frac{x}{2\sqrt{x}\left(1+3x^2 \right)}=-\frac{\sqrt{x}}{2\left(1+3x^2 \right)}$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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