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:

$$f(x)=0-\frac{\left(\sqrt{x} \right)^2}{1+3\left(\sqrt{x} \right)^4}\frac{d}{dx}\left(\sqrt{x} \right)=-\frac{x}{2\sqrt{x}\left(1+3x^2 \right)}=-\frac{\sqrt{x}}{2\left(1+3x^2 \right)}$$
 
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