# PARTIAL DIFFERENTIATION (Text)

$\color {royalblue} {Definition}:\ \hspace{20cm}$
$Let\ u\ =\ f ( x , y )\ then\ the\ partial\ differentiation\ of\ u\ with\ respect\ to\ x\ is\ defined\ as$
$differentiation\ of\ u\ w.\ r.\ t\ x\ treating\ y\ as\ constant\ and\ is\ denoted\ by\ \frac{∂u}{∂x}$
$Similarly\ partial\ differentiation\ of\ u\ with\ respect\ to\ y\ is\ defined\ as$
$differentiation\ of\ u\ w.\ r.\ t\ y\ treating\ x\ as\ constant\ and\ is\ denoted\ by\ \frac{∂u}{∂y}$
$\color {royalblue} {Problems}:\ \hspace{20cm}$
$\color {purple} {Example\ 1:}\ \color {red} {Find\ \frac{∂u}{∂x}\ and\ \frac{∂u}{∂y}}\ if\ u\ = x^2\ +\ y^2\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^2\ +\ y^2\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^2)\ +\ \frac{∂}{∂x}( y^2)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 2\ x\ +\ 0\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 2\ x\ \hspace{10cm}$
$u\ = x^2\ +\ y^2\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ \frac{∂}{∂y}( x^2)\ +\ \frac{∂}{∂y}( y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 2\ y\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 2\ y\ \hspace{10cm}$
$\color {purple} {Example\ 2:}\ If\ u\ = x^3\ +\ y^3\ +\ 3\ x^2\ y\ +\ 3\ x\ y^2,\ \color {red} {find\ \frac{∂u}{∂x}\ and\ \frac{∂u}{∂y}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^3\ +\ y^3\ +\ 3\ x^2\ y\ +\ 3\ x\ y^2\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}( y^3)\ +\ 3\ \frac{∂}{∂x}( x^2)\ y\ +\ 3\ y^2\ \frac{∂}{∂x}(x)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ +\ 3\ (2\ x)\ y)\ +\ 3\ y^2\ (1)\ \frac{∂}{∂x}( y))\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 6\ x\ y\ +\ 3\ y^2\ \hspace{10cm}$
$u\ = x^3\ +\ y^3\ +\ 3\ x^2\ y\ +\ 3\ x\ y^2\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ \frac{∂}{∂y}( x^3)\ +\ \frac{∂}{∂y}( y^3)\ +\ 3\ x^2\ \frac{∂}{∂y}(y)\ +\ 3\ x\ \frac{∂}{∂y}(y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ +\ 3\ x^2\ (1)\ +\ 3\ x\ (2\ y)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ y^2\ +\ 3\ x^2\ +\ 6\ x\ y\ \hspace{10cm}$
$\color {purple} {Example\ 3:}\ If\ u\ = 2\ x^3\ +\ 4\ \ y^3\ +\ 2\ x\ y\ ,\ \color {red} {find\ \frac{∂u}{∂x}\ and\ \frac{∂u}{∂y}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = 2\ x^3\ +\ 4\ \ y^3\ +\ 2\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ 2\ \frac{∂}{∂x}( x^3)\ +\ 4\ \frac{∂}{∂x}( y^3)\ +\ 2\ y\ \frac{∂}{∂x}( x)\ \ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 2\ (3\ x^2)\ +\ 0\ +\ 2\ y\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 6\ x^2\ +\ 2\ y\ \hspace{10cm}$
$u\ = 2\ x^3\ +\ 4\ \ y^3\ +\ 2\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ 2\ \frac{∂}{∂y}( x^3)\ +\ 4\ \frac{∂}{∂y}( y^3)\ +\ 2\ x\ \frac{∂}{∂y}( y)\ \ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 2\ (0)\ +\ 4(3\ y^2)\ +\ 2\ x\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 12\ y^2\ +\ 2\ x\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 12\ y^2\ +\ 2\ x\ \hspace{10cm}$

$\color {purple} {Example\ 4:}\ If\ u\ =\ log(x^2\ +\ y^2)\ ,\ \color {red} {find\ \frac{∂u}{∂x}\ ,\ \frac{∂u}{∂y}\ and\ \frac{∂^2u}{∂y\ ∂x}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ =\ log(x^2\ +\ y^2)\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}(log(x^2\ +\ y^2))\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{1}{x^2\ +\ y^2}\ \frac{∂}{∂x}(x^2\ +\ y^2)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{1}{x^2\ +\ y^2}\ (2\ x\ +\ 0)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{2\ x}{x^2\ +\ y^2}\ \hspace{10cm}$
$u\ =\ log(x^2\ +\ y^2)\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ \frac{∂}{∂y}(log(x^2\ +\ y^2))\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{1}{x^2\ +\ y^2}\ \frac{∂}{∂y}(x^2\ +\ y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{1}{x^2\ +\ y^2}\ (0\ +\ 2\ y)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{2\ y}{x^2\ +\ y^2}\ \hspace{10cm}$
$\frac{∂^2u}{∂y\ ∂x}\ =\ \frac{∂}{∂y}(\frac{2\ x}{x^2\ +\ y^2})\ \hspace{10cm}$
$\frac{∂^2u}{∂y\ ∂x}\ =\ \frac{(x^2\ +\ y^2)\ \frac{∂}{∂y}(2\ x)\ -\ 2\ x\ \frac{∂}{∂y}(x^2\ +\ y^2)}{(x^2\ +\ y^2)^2}\ \hspace{10cm}$
$\frac{∂^2u}{∂y\ ∂x}\ =\ \frac{(x^2\ +\ y^2)\ (0)\ -\ 2\ x\ (0\ +\ 2\ y)}{(x^2\ +\ y^2)^2}\ \hspace{10cm}$
$\frac{∂^2u}{∂y\ ∂x}\ =\ \frac{-\ 4\ x\ y}{(x^2\ +\ y^2)^2}\ \hspace{10cm}$
$\color {purple} {Example\ 5:}\ If\ u\ =\ 2\ x^3\ -\ 3\ x^2y\ +\ 3\ x y^2\ +\ 5\ y^3,\ \color {red} {find\ the\ value\ of\ \ x\ \frac{∂u}{∂x}\ +\ y\ \frac{∂u}{∂y}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = 2\ x^3\ -\ 3\ x^2y\ +\ 3\ x y^2\ +\ 5\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ 2\ \frac{∂}{∂x}( x^3)\ -\ 3\ y\ \frac{∂}{∂x}( x^2)\ +\ 3\ y^2\ \frac{∂}{∂x}( x)\ +\ 5\ \frac{∂}{∂x}(y^3)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 2\ (3\ x^2)\ -\ 3y\ (2x)\ +\ 3\ y^2\ (1)\ +\ 5(0)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 6\ x^2\ -\ 6\ xy\ +\ 3\ y^2\ \hspace{10cm}$
$u\ = 2\ x^3\ -\ 3\ x^2y\ +\ 3\ x y^2\ +\ 5\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ 2\ \frac{∂}{∂y}( x^3)\ -\ 3\ x^2\ \frac{∂}{∂y}(y)\ +\ 3\ x\ \frac{∂}{∂y}(y^2)\ +\ 5\ \frac{∂}{∂y}(y^3)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 2\ (0)\ -\ 3\ x^2\ (1)\ +\ 3\ x\ (2y)\ +\ 5(3y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ -\ 3\ x^2\ +\ 6\ xy\ +\ 15\ y^2\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ x (6\ x^2\ -\ 6\ xy\ +\ 3\ y^2)\ +\ y(-\ 3\ x^2\ +\ 6\ xy\ +\ 15\ y^2)\ \hspace{10cm}$
$=\ 6\ x^3\ -\ 6\ x^2\ y\ +\ 3\ x\ y^2\ -\ 3\ x^2\ y\ +\ 6\ x\ y^2\ +\ 15\ y^3\ \hspace{10cm}$
$=\ 6\ x^3\ -\ 9\ x^2\ y\ +\ 9\ x\ y^2\ +\ 15\ y^3\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ 6\ x^3\ -\ 9\ x^2\ y\ +\ 9\ x\ y^2\ +\ 15\ y^3\ \hspace{10cm}$

$\color {purple} {Example\ 6:}\ If\ u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ ,\ \color {red} {prove\ that\ x\ \frac{∂u}{∂x}\ +\ y\ \frac{∂u}{∂y}\ =\ 3\ u}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ ——–\ (1) \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ ‘\ x\ ‘\ \ on\ both\ sides\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{∂}{∂x}(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}( y^3)\ +\ 3\ y^2\ \frac{∂}{∂x}( x)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ + 3\ y^2\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ + 3\ y^2\ ——\ (2)\ \hspace{10cm}$
$u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ ‘\ y\ ‘\ \ on\ both\ sides\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{∂}{∂y}(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{∂}{∂y}( x^3)\ +\ \frac{∂}{∂y}( y^3)\ +\ 3\ x\ \frac{∂}{∂y}(y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ + 3\ x\ (2y)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ y^2\ + 6\ x\ y\ ——\ (3)\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ x (3\ x^2\ + 3\ y^2)\ +\ y(3\ y^2\ + 6\ x\ y)\ \hspace{10cm}$
$=\ 3\ x^3\ +\ 3\ x\ y^2\ +\ 3\ y^3\ +\ 6\ x\ y^2\ \hspace{10cm}$
$=\ 3\ x^3\ +\ 3\ y^3\ +\ 9\ x\ y^2\ \hspace{10cm}$
$=\ 3(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}$
$=\ 3\ (u)\ —– using (1)\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ 3\ u\ \hspace{10cm}$
$\color {purple} {Example\ 7:}\ If\ u\ = \frac{x^3\ y^3}{x^3\ +\ y^3} ,\ \color {red} {Show\ that\ x\ \frac{∂u}{∂x}\ +\ y\ \frac{∂u}{∂y}\ =\ 3\ u}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ u\ =\ \frac{x^3\ y^3}{x^3\ +\ y^3}\ ——–\ (1) \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ ‘\ x\ ‘\ \ on\ both\ sides\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{∂}{∂x}(\frac{x^3\ y^3}{x^3\ +\ y^3})\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{(x^3\ +\ y^3)\ \frac{∂}{∂x}(x^3\ y^3)\ -\ x^3\ y^3\ \frac{∂}{∂x}(x^3\ +\ y^3)}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{(x^3\ +\ y^3)\ (y^3\ 3\ x^2)\ -\ x^3\ y^3\ (3\ x^2\ +\ 0)}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{3\ x^5\ y^3\ +\ 3\ x^2\ y^6\ -\ 3\ x^5\ y^3}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ \frac{3\ x^2\ y^6}{(x^3\ +\ y^3)^2}\ ——\ (2)\ \hspace{10cm}$
$u\ =\ \frac{x^3\ y^3}{x^3\ +\ y^3}\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ ‘\ y\ ‘\ \ on\ both\ sides\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{∂}{∂y}(\frac{x^3\ y^3}{x^3\ +\ y^3})\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{(x^3\ +\ y^3)\ \frac{∂}{∂y}(x^3\ y^3)\ -\ x^3\ y^3\ \frac{∂}{∂y}(x^3\ +\ y^3)}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{(x^3\ +\ y^3)\ (x^3\ 3\ y^2)\ -\ x^3\ y^3\ (0\ +\ 3\ y^2)}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{3\ x^6\ y^2\ +\ 3\ x^3\ y^5\ -\ 3\ x^3\ y^5}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ \frac{3\ x^6\ y^2}{(x^3\ +\ y^3)^2}\ ——\ (3)\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ x (\frac{3\ x^2\ y^6}{(x^3\ +\ y^3)^2})\ +\ y(\frac{3\ x^6\ y^2}{(x^3\ +\ y^3)^2})\ \hspace{10cm}$
$=\ \frac{3\ x^3\ y^6}{(x^3\ +\ y^3)^2}\ +\ \frac{3\ x^6\ y^3}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$=\ \frac{3\ x^3\ y^6\ +\ 3\ x^6\ y^3}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$=\ \frac{3\ x^3\ y^3(y^3\ +\ x^3)}{(x^3\ +\ y^3)^2}\ \hspace{10cm}$
$=\ \frac{3\ x^3\ y^3}{x^3\ +\ y^3}\ \hspace{10cm}$
$=\ 3\ (u)\ —– using (1)\ \hspace{10cm}$
$x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ 3\ u\ \hspace{10cm}$
$\color {purple} {Example\ 8:}\ If\ u\ =\ x^3\ -\ 2\ x^2\ y\ +\ 3\ x\ y^2\ +\ y^3\ ,\ \color {red} {find\ \frac{∂^2u}{∂x^2}\ and\ \frac{∂^2u}{∂y^2}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^3\ -\ 2\ x^2\ y\ +\ 3\ x\ y^2\ +\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ -\ 2\ y\ \frac{∂}{∂x}( x^2)\ +\ 3\ y^2\ \frac{∂}{∂x}( x)\ +\ \frac{∂}{∂x}(y^3)\ \ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ -\ 2\ y\ (2\ x)\ +\ 3\ y^2\ (1)\ +\ 0\ \frac{∂}{∂x}( y)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ -\ 4\ x\ y\ +\ 3\ y^2\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2\ -\ 4\ x\ y\ +\ 3\ y^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ -\ 4\ y\ \frac{∂}{∂x}( x)\ +\ \frac{∂}{∂x}(3\ y^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ -\ 4\ y\ (1)\ +\ 0\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ -\ 4\ y\ \hspace{10cm}$
$u\ = x^3\ -\ 2\ x^2\ y\ +\ 3\ x\ y^2\ +\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ \frac{∂}{∂y}( x^3)\ -\ 2\ x^2\ \frac{∂}{∂y}( y)\ +\ 3\ x\ \frac{∂}{∂y}(y^2)\ +\ \frac{∂}{∂y}(y^3)\ \ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ -\ 2\ x^2\ (1)\ +\ 3\ x\ (2\ y)\ +\ 3\ y^2\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ -\ 2\ x^2\ +\ 6\ x\ y\ +\ 3\ y^2\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}(\frac{∂u}{∂y})\ =\ \frac{∂}{∂y}(-\ 2\ x^2\ +\ 6\ x\ y\ +\ 3\ y^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ \frac{∂}{∂y}( -2\ x^2)\ +\ 6\ x\ \frac{∂}{∂y}(y)\ +\ 3\ \frac{∂}{∂y}(y^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ 0\ +\ 6\ x\ (1)\ +\ 3(2\ y)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ 6\ x\ +\ 6\ y\ \hspace{10cm}$
$\color {purple} {Example\ 9:}\ If\ u\ =\ x^3\ +\ y^3\ +\ 4\ x\ y,\ \color {red} {find\ the\ x^2\ \frac{∂^2u}{∂x^2}\ +\ y^2\ \frac{∂^2u}{∂y^2}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^3\ +\ y^3\ +\ 4\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}(y^3)\ +\ 4\ y\ \frac{∂}{∂x}( x)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ +\ 4\ y (1)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 4\ y\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2\ +\ 4\ y)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ +\ 4 \frac{∂}{∂x}(4\ y)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ +\ 0\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ \hspace{10cm}$
$u\ =\ x^3\ +\ y^3\ +\ 4\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ +\ 4\ x( 1)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ y^2\ +\ 4\ x\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}(\frac{∂u}{∂y})\ =\ \frac{∂}{∂y}(3\ y^2\ +\ 4\ x)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ 3\ \frac{∂}{∂y}( y^2)\ +\ \frac{∂}{∂y}( 4\ x)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ 3(2y)\ +\ 0\ \hspace{10cm}$
$\frac{∂^2\ u}{∂y^2}\ =\ 6\ y\ \hspace{10cm}$
$x^2\ \frac{∂^2\ u}{∂x^2}\ +y\ \frac{∂^2\ u}{∂y^2}\ =\ x^2 (6\ x)\ +\ y^2(y)\ \hspace{10cm}$
$=\ 6\ x^3\ +\ 6\ y^3\ \hspace{10cm}$