Not exactly Partial Derivative I've been just introduced to concept of Partial Derivative, My question is
for some continuous and differentiable $g(x)$ we have
$$g'(x)=\lim\limits_{\delta x \to 0}  \left( \frac{g(x+\delta x) - g(x)}{\delta x} \right)$$
similarly for , for some continuous and differentiable $f(x,y,z)$ we have
$$\frac {\partial f}{\partial y}=\lim\limits_{\delta y \to 0}  \left( \frac{f(x,y+\delta y,z) - f(x,y,z)}{\delta y} \right)$$
That means I'm looking at change in $f(x,y,z)$ w.r.t $y$, keeping all other variables constant
But What if I am not interested in observing change in $f(x,y,z)$ w.r.t. one variable keeping all other constant.
Actually, I want to observe change in $f(x,y,z)$ w.r.t.  both $x$ & $y$ at once keeping $z$ constant.
will I get what I'm looking for ( i.e. change in $f(x,y,z)$ w.r.t. change in $x$ & $y$ ) by evaluating following two expressions?
$$\frac{\partial f^2 }{\partial y \partial x} \tag{1}$$
$$\frac{\partial f^2 }{\partial x \partial y } \tag{2}$$
In that case both $1$ & $2$ should be equal. And
If both
$$\frac{\partial f }{\partial y}   
 \& \frac{\partial f }{\partial x } $$
are continuous and differentiable then
$$\frac{\partial f^2 }{\partial x \partial y } = \frac{\partial f^2 }{\partial y \partial x } =\lim\limits_{ \delta y \to 0 , \delta x \to 0}  \left( \frac{f(x+\delta x ,y+ \delta y,z) - f(x,y,z)}{\delta x  + \delta y} \right)$$
Is above stated expression mathematically correct ( I don't think it could be, cause I myself proposed it ), If so,
How can we prove all the equalities mentioned in it If, not what is correct expression in form of limits
 A: What you are looking for are directional derivatives. You can make the derivative in the dirction of the unit vector $\vec u$. This is the definition: $$
\nabla_\vec{u}f(\vec x)= \lim_{h \rightarrow 0}{\frac{f(\vec{x} + h\vec{u}) - f(\vec{x})}{h}}$$
If f is differentiable:$$
\nabla_\vec{u}f(\vec x)=\nabla f(\vec x) \cdot \vec u
$$
Where $\nabla f(x)$ is the gradient of f (the vector of its partial derivatives):
$$
\nabla f(\vec x)=(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots,\frac{\partial f}{\partial x_n})
$$
You could set $\vec u = \frac{1}{\sqrt 2}(1,1,0)$ and calculate $\nabla_\vec{u}f(\vec x)$.
A: The derivative expresses that variations of a function are approximately proportional to variations in the arguments, for small variations.
In other words, $\delta f(x)\approx\lambda\delta x$, for some $\lambda$, that is called the derivative and is denoted by $f'(x)$.
$$f(x+\delta x)\approx f(x)+f'(x)\delta x.$$
When there are other variables held constants, the partial derivative notation makes it explicit:
$$f(x+\delta x,y,z)\approx f(x,y,z)+\frac{\partial f}{\partial x}\delta x.$$
You can vary two variables at a time if you want, and linearity applies, so that you can write
$$f(x+\delta x,y+\delta y,z)\approx f(x,y,z)+\frac{\partial f}{\partial x}\delta x+\frac{\partial f}{\partial y}\delta y.$$
No second order derivatives are involved.
