# PARTIAL DIFFERENTIATION (Excercise)

$\LARGE{\color {purple} {PART- A}}$
$1.\ If\ u\ =\ x^3\ +\ y^3\ ,\ find\ \frac{∂^2u}{∂x^2}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ u\ =\ x^3\ +\ y^3\ \hspace{15cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}(y^3)\ \ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ \frac{∂}{∂x}( y)\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- B}}$
$2.\ If\ u\ = 3\ x^3\ +\ 4\ \ y^3\ +\ 6\ x\ y\ ,\ find\ (i)\ \frac{∂u}{∂x}\ (ii)\ \frac{∂u}{∂y}\ \hspace{15cm}$
$\color {black}{Solution:}\ (i)\ Given\ u\ = 3\ x^3\ +\ 4\ \ y^3\ +\ 6\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ 3\ \frac{∂}{∂x}( x^3)\ +\ 4\ \frac{∂}{∂x}( y^3)\ +\ 6\ y\ \frac{∂}{∂x}( x)\ \ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ (3\ x^2)\ +\ 0\ +\ 6\ y\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 9\ x^2\ +\ 6\ y\ \hspace{10cm}$
$(ii)\ u\ = 3\ x^3\ +\ 4\ \ y^3\ +\ 6\ x\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ 3\ \frac{∂}{∂y}( x^3)\ +\ 4\ \frac{∂}{∂y}( y^3)\ +\ 6\ x\ \frac{∂}{∂y}( y)\ \ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ (0)\ +\ 4(3\ y^2)\ +\ 6\ x\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 12\ y^2\ +\ 6\ x\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 12\ y^2\ +\ 6\ x\ \hspace{10cm}$
$\color {purple} {3:}\ If\ u\ = x^4\ +\ y^3\ +\ 2\ x^2\ y^2\ +\ 3\ x^2\ y,\ \color {red} {find\ \frac{∂u}{∂x}\ and\ \frac{∂u}{∂y}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^4\ +\ y^3\ +\ 2\ x^2\ y^2\ +\ 3\ x^2\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^4)\ +\ \frac{∂}{∂x}( y^3)\ +\ 2\ y^2\frac{∂}{∂x}( x^2)\ \ +\ 3\ y\ \frac{∂}{∂x}(x^2)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 4\ x^3\ +\ 0\ +\ 2\ y^2\ (2\ x)\ +\ 3\ y\ (2x)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 4\ x^3\ +\ 4\ x\ y^2\ +\ 6\ xy\ \hspace{10cm}$
$u\ = x^4\ +\ y^3\ +\ 2\ x^2\ y^2\ +\ 3\ x^2\ y\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ \frac{∂}{∂y}( x^4)\ +\ \frac{∂}{∂y}( y^3)\ +\ 2\ x^2\ \frac{∂}{∂y}(y^2)\ +\ 3\ x^2\ \frac{∂}{∂y}(y)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ +\ 2\ x^2\ (2y)\ +\ 3\ x^2\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ y^2\ +\ 4\ x^2\ y\ +\ 3\ x^2\ \hspace{10cm}$
$\color {purple} {4:}\ If\ u\ =\ x^3\ +\ y^3\ +\ 3\ xy\ ,\ \color {red} {find\ \frac{∂^2u}{∂x^2}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = x^3\ +\ y^3\ +\ 3\ xy\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}( y^3)\ +\ 3\ y\ \frac{∂}{∂x}( x)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ +\ 3\ y\ (1)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 3\ y\ \hspace{10cm}$
$Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2\ +\ 3\ y)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ +\ 3\ \frac{∂}{∂x}(y)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ +\ 3(0)\ \hspace{10cm}$
$\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {5:}\ If\ u\ = 2\ x^3\ +\ 3\ x^2y\ +\ 4\ x y^2\ +\ 4\ y^3,\ \color {red} {find\ \frac{∂u}{∂x}\ and\ \frac{∂u}{∂y}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ u\ = 2\ x^3\ +\ 3\ x^2y\ +\ 4\ x y^2\ +\ 4\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂x}\ (u)\ =\ 2\ \frac{∂}{∂x}( x^3)\ +\ 3\ y\ \frac{∂}{∂x}( x^2)\ +\ 4\ y^2\ \frac{∂}{∂x}( x)\ +\ 4\ \frac{∂}{∂x}(y^3)\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 2\ (3\ x^2)\ +\ 3y\ (2x)\ +\ 4\ y^2\ (1)\ +\ 0\ \hspace{10cm}$
$\frac{∂u}{∂x}\ =\ 6\ x^2\ +\ 6\ xy\ +\ 4\ y^2\ \hspace{10cm}$
$u\ = 2\ x^3\ +\ 3\ x^2y\ +\ 4\ x y^2\ +\ 4\ y^3\ \hspace{15cm}$
$Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}$
$\frac{∂}{∂y}\ (u)\ =\ 2\ \frac{∂}{∂y}( x^3)\ +\ 3\ x^2\ \frac{∂}{∂y}(y)\ +\ 4\ x\ \frac{∂}{∂y}(y^2)\ +\ 4\ \frac{∂}{∂y}(y^3)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 2\ (0)\ +\ 3\ x^2\ (1)\ +\ 4\ x\ (2y)\ +\ 4(3y^2)\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 0\ +\ 3\ x^2\ +\ 8\ xy\ +\ 12\ y^2\ \hspace{10cm}$
$\frac{∂u}{∂y}\ =\ 3\ x^2\ +\ 8\ xy\ +\ 12\ y^2\ \hspace{10cm}$

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