Partial derivative difference question

tony873004
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What's the difference between \partial ^2 x and \partial x^2?

Is \partial ^2 x the same as \left( {\partial x} \right)^2 like \sin ^2 x$ is the same as \left( {\sin x} \right)^2?

Thanks!
 
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If you have \frac{\partial^2 x}{\partial t^2}, this translates to \frac{\partial}{\partial t} \frac{\partial x}{\partial t}. In other words, in derivative notation, it's common to write \partial t * \partial t as \partial t^2. Does that help?
 
so does it mean the second partial derivative, rather than squaring something?
 
In the case of \frac{\partial^2 x}{\partial t^2}, it means the second partial derivative of a function x with respect to t. You can think of the "partial" terms as being squared, but I'm not sure how that would be of any help unless you're having to separate something like \frac{\partial^2 x}{\partial t^2}, as was done above. Is there a particular problem that's troubling you?
 
hotcommodity said:
Is there a particular problem that's troubling you?
Yes, this is a small part of a larger problem. But I think I can get it now that I understand the notation. If I get stuck, I'll post the entire problem. Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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