PARTIAL DIFFERENTIATION (Excercise)

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