Derivative of Integral: Does the $x^2$ Cancel?

[karush]

https://www.physicsforums.com/attachments/5396
I assume the derivative cancels the intregal but the $x^2$ ?

 

[MarkFL]

The Leibniz Integral Rule states:

\(\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f(h(x))h'(x)-f(g(x))g'(x)\)

Can you proceed?

 

[karush]

$h\left(x\right)=10h'\implies\left(x\right)=0$
$g\left(x\right)={x}^{2} \implies g'\left(x\right)=2x$
$f\left(z\right)=\frac{1}{{z}^{2}}+1$

I think that is $x^2$ it a fuzzy pic

I have to submit to preview the site freezes up if I preview on the edit page

 

[MarkFL]

The image is hard to read, but it appear to be that we are asked to evaluate:

\(\displaystyle \frac{d}{dx}\int_{x^3}^{10}\frac{dz}{z^2+1}\)

And if this is the case, then we have:

\(\displaystyle g(x)=x^3\implies g'(x)=3x^2\)

\(\displaystyle h(x)=10\implies h'(x)=0\)

\(\displaystyle f(z)=\frac{1}{z^2+1}\)

 

[karush]

The original pic looks like $x^2$

So I got
$\frac{2x}{{x}^{4}+1}$

I hope

 

[MarkFL]

If the lower limit is $x^2$, then we have:

\(\displaystyle \frac{d}{dx}\int_{x^2}^{10}\frac{dz}{z^2+1}=\frac{1}{10^2+1}\cdot0-\frac{1}{(x^2)^2+1}\cdot2x=-\frac{2x}{x^4+1}\)

 

[karush]

Cool, always learn a lot from MHB