Derivative of T(x, y, z): Get the Answer Now!

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Homework Statement



T(x, y, z) = 200e^(−x^2−3y^2−9z^2 )
I'm not sure how to get the derivative in terms of x,y or z.



The Attempt at a Solution



For x: -200x^2e^(−x^2−3y^2−9z^2 )(-2x)? Is this right? Probably not, how do i do it?
 
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killersanta said:

Homework Statement



T(x, y, z) = 200e^(−x^2−3y^2−9z^2 )
I'm not sure how to get the derivative in terms of x,y or z.



The Attempt at a Solution



For x: -200x^2e^(−x^2−3y^2−9z^2 )(-2x)? Is this right? Probably not, how do i do it?
This is not quite the correct partial derivative of T with respect to x. That's the terminology that is usually used.

The corrected version is -200e^(−x^2−3y^2−9z^2 )(-2x)

The notation can appear in two forms:
\frac{\partial T}{\partial x}
or
Tx
 
thank you!
 
Do you understand why that x2 factor you had shouldn't be there?
 
Mark44 said:
Do you understand why that x2 factor you had shouldn't be there?

No, I thought you pulled it down? Why?
 
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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