$\displaystyle \[\frac{d}{{dx}}\left( {\,{x^{1/2}}\int {{x^{3/2}}f(x)dx} } \right)\]$

????

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- Jun 23rd 2013, 08:29 AMmpx86derivative problem 2
$\displaystyle \[\frac{d}{{dx}}\left( {\,{x^{1/2}}\int {{x^{3/2}}f(x)dx} } \right)\]$

???? - Jun 23rd 2013, 11:23 PMchiroRe: derivative problem
Hey mpx86.

Hint: Use the product rule for differentiation and also use the chain rule for the fundamental theorem of calculus. - Jun 24th 2013, 10:05 AMhollywoodRe: derivative problem
I see how to apply the product rule and I can use the fundamental theorem of calculus on one of the terms. The other term becomes $\displaystyle \frac{1}{2x^{1/2}}\int{x^{3/2}f(x)\,dx$. Can it be simplified from there?

- Hollywood - Jun 24th 2013, 04:24 PMchiroRe: derivative problem
I don't think that you can simplify in general any further.

- Jun 24th 2013, 04:49 PMadkinsjrRe: derivative problem
I have to ask about this one myself.

After the product rule I have:

$\displaystyle \frac{1}{2\sqrt{x}}\int x^{\frac{3}{2}}f(x)dx + \sqrt{x}\frac{d}{dx}\int x^{\frac{3}{2}}f(x)dx$

I don't see how you would differentiate the integral in the second term with the FTOC because it is indefinite, the FTOC is always stated in terms of definite integrals as far as I understand it. I guess if say that $\displaystyle \frac{d}{dx}\int g(x)dx=\frac{d}{dx}(G(x)+C)=g(x)$ is always true so I can say my $\displaystyle x^{\frac{3}{2}}f(x)$ is my $\displaystyle g(x)$ and just say $\displaystyle \frac{d}{dx}\int x^{\frac{3}{2}}f(x)dx= x^{\frac{3}{2}}f(x)$

Or maybe I shouldn't ask if it is true, it is just definition of integral? Where does chain rule come in? - Jun 24th 2013, 04:56 PMchiroRe: derivative problem
For some reason, I thought that the integral had limits that were a function of x.

The post has been edited and I don't know if this was changed or not. - Jun 24th 2013, 08:01 PMmpx86Re: derivative problem