What is a total differential? (Geometrically speaking?)

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Just got around to learning total differentials in calculus 3. Or total derivative, having a hard time wrapping my mind around what it is geometrically speaking. Can someone explain?
 
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MathWarrior said:
Just got around to learning total differentials in calculus 3. Or total derivative, having a hard time wrapping my mind around what it is geometrically speaking. Can someone explain?

Lets take a single variable case here, say y=f(x).

So the total differential of f is df = f'(x)dx. You can think of df as a function of two variables, x and dx ( Where dx is also independent here of course ).

Think of an infinitesimally small quantity when you think about dx. So for very very small values of dx you get :

f(x, dx) ≈ f'(x) ( Roughly equal to ).

In a way we cheat the system when we take the limit as h→0 of the definition of the limit. Remember how h is close to 0, but not zero exactly? Think of dx in the same manner. Extremely small, but not zero exactly.
 
Let f be a differentiable function in x, y, and z (which is stronger than just saying that the partial derivatives exist). Further, suppose x, y, and z are differentiable functions of some parameter t. Then f(t)= f(x(t), y(t), z(t)) describes f values along a specific smooth path in R3. If you like, you can think of it as the trajectory of an object in space.

By the chain rule,
\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{df}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}

Since that is an ordinary derivative of a function of a single variable, we can define its "differential":
df= \frac{df}{dt} dt= \left(\frac{\partial f}{\partial x}\frac{df}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}\right)dt

Or course, that is the same as
df= \frac{df}{dt} dt= \frac{\partial f}{\partial x}\frac{df}{dt}dt+ \frac{\partial f}{\partial y}\frac{dy}{dt}dt+ \frac{\partial f}{\partial z}\frac{dz}{dt}dt

df= \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz
where all mention of "t" has disappeared.
 
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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