Q.If $V(x,y)=e^x(x\cos y-y\sin y)$, then prove that $\dfrac{\partial^2 V}{\partial x^2}+\dfrac{\partial^2 V}{\partial y^2}=0$.

For a function of one variable, the derivative measures the rate of change of $f$ with respect to $x$. For $F(x,y)$, it is meaningful to ask for the rate of change with respect to $x$ alone, holding $y$ fixed — this is a partial derivative.

Definition. Let $A=\{(x,y)\mid a<x<b,\,c<y<d\}\subset\mathbb R^2$, $F:A\to\mathbb R$, $(x_0,y_0)\in A$.

1. $F$ has a partial derivative w.r.t. $x$ at $(x_0,y_0)$ if $\displaystyle\lim_{h\to0}\frac{F(x_0+h,y_0)-F(x_0,y_0)}{h}$ exists; the limit value is $\dfrac{\partial F}{\partial x}(x_0,y_0)$, also written $F_x(x_0,y_0)$.
2. $F$ has a partial derivative w.r.t. $y$ at $(x_0,y_0)$ if $\displaystyle\lim_{k\to0}\frac{F(x_0,y_0+k)-F(x_0,y_0)}{k}$ exists; the limit value is $\dfrac{\partial F}{\partial y}(x_0,y_0)$, also written $F_y(x_0,y_0)$. Read $\partial F/\partial x$ as "partial $F$ by partial $x$" (or "dho $F$ by dho $x$"). Geometrically, $F(x,y_0)$ is the curve obtained by slicing the surface $z=F(x,y)$ with the plane $y=y_0$, and $\dfrac{\partial F}{\partial x}(x_0,y_0)$ is the slope of the tangent to that curve at $x=x_0$; symmetrically for $\dfrac{\partial F}{\partial y}$.
   Tip

   Mechanically, computing $\dfrac{\partial F}{\partial x}$ just means differentiating $F$ with respect to $x$ by the ordinary rules, treating every OTHER variable as though it were a constant. All the familiar rules — sum, product, quotient, chain rule — carry over unchanged; only the "which letters are constants" bookkeeping is new. Partial derivatives extend to three or more variables the same way, holding all other variables fixed in turn.

   Watch out

   Existence of $F_x$ and $F_y$ at a point does not imply $F$ is continuous there (unlike the one-variable case, where differentiability always implies continuity). The classic example: $f(x,y)=0$ if $xy\ne0$ and $f(x,y)=1$ if $xy=0$. Both $f_x(0,0)$ and $f_y(0,0)$ exist and equal $0$ (moving along either axis keeps $f\equiv1$), yet $f$ is not even continuous at $(0,0)$ (along $y=x,\,x\ne0$, $f\to0\ne f(0,0)=1$).

   Higher-order and mixed partial derivatives. Since $F_x$ is again a function of $(x,y)$, it can itself be partially differentiated: …