An attempt to find the total differential of a two-variable function

[Kashmir]

Let $\quad z=h(x, y)$
and
$x=f(t) ; y=g(t)$

Let the change in the function z be given by $\Delta z=h(x+\Delta x, y+\Delta y)-h(x,y)$

We can also write the change as

$\begin{aligned} \Delta z=h &(x+\Delta x, y)-\\ & h(x, y)-h(x+\Delta x, y) \\ &+h(x+\Delta x, y+\Delta y) \end{aligned}$$\Delta z=\Delta h_{y\text { constant }}+\Delta h_{x \operatorname{constant} }$

In the limit then we have
$dz=\frac{\partial h}{\partial x} d x+\frac{\partial h}{\partial y} d y$

Is there anything wrong with this derivation?

 

[anuttarasammyak]

Sounds good.

 

[BvU]

Is there anything wrong with this derivation?

What could possibly be wrong ?

$$
\Delta z=\Delta h_{y\text { constant }}+\Delta h_{x

\operatorname{constant} }$$is not really clear

 

[Kashmir]

I've not seen this derivation, thought maybe it was wrong somehow.

What could possibly be wrong ?

$$
\Delta z=\Delta h_{y\text { constant }}+\Delta h_{x

\operatorname{constant} }$$is not really clear

The delta y constant means that y has been kept as a constant

 

[anuttarasammyak]

Some amendment
\triangle z = \triangle h_{y\ constant}+\triangle h_{x+\triangle x \ constant}
\triangle z = \triangle h_{x\ constant}+\triangle h_{y+\triangle y \ constant}
For general paths
\triangle z = \int _{(x,y)}^{(x+\triangle x,y+\triangle y)} \nabla h \cdot \mathbf{dl}

 

[Kashmir]

Some amendment
\triangle z = \triangle h_{y\ constant}+\triangle h_{x+\triangle x \ constant}
\triangle z = \triangle h_{x\ constant}+\triangle h_{y+\triangle y \ constant}
For general paths
\triangle z = \int _{(x,y)}^{(x+\triangle x,y+\triangle y)} \nabla h \cdot \mathbf{dl}

So am i wrong, friend?

 

[anuttarasammyak]

The difference is magnitude of second order of infinitesimals so

In the limit then we have
dz=∂h∂xdx+∂h∂ydy

holds in the limit.