Question

Please refer to the pic

Answer 1

Step-by-step explanation:

We have,

$u = {x}^{y}$

$\implies \: \frac{ \partial{u}}{ \partial{x}} = y .{x}^{y - 1}$

$\implies \: \frac{ \partial}{ \partial{x}} \bigg( \frac{ \partial{u}}{ \partial{x}} \bigg) = y(y - 1) .{x}^{y - 2}$

$\implies \: \frac{ \partial^{2} {u}}{ \partial{x} ^{2} } = y(y - 1) .{x}^{y - 2}$

$\implies \: \frac{ \partial}{ \partial{y}} \bigg(\frac{ \partial^{2} {u}}{ \partial{x} ^{2} } \bigg) = y(y - 1) .{x}^{y - 2}$

$\implies \: \frac{ \partial}{ \partial{y}} \bigg(\frac{ \partial^{2} {u}}{ \partial{x} ^{2} } \bigg) = (y - 1) .{x}^{y - 2} + y. {x}^{y - 2} + y(y - 1) {x}^{y - 2}. \ln(x)$

$\implies \: \frac{ \partial^{3} {u}}{ \partial{x} ^{2} \partial{y} } = \{(y - 1) + y + y(y - 1). \ln(x) \}. {x}^{y - 2}$

$\implies \: \frac{ \partial^{3} {u}}{ \partial{x} ^{2} \partial{y} } = \{y - 1 + y + ( {y}^{2} - y). \ln(x) \}. {x}^{y - 2}$

$\implies \: \frac{ \partial^{3} {u}}{ \partial{x} ^{2} \partial{y} } = \{2y - 1 + ( {y}^{2} - y). \ln(x) \}. {x}^{y - 2}$

Now,

$\implies \: \frac{ \partial{u}}{ \partial{x}} = y .{x}^{y - 1}$

$\implies \: \frac{ \partial}{ \partial{y}} \bigg( \frac{ \partial{u}}{ \partial{x}} \bigg) = {x}^{y - 1} + y .{x}^{y - 1} . \ln(x)$

$\implies \: \frac{ \partial^{2} {u} }{ \partial{x} \partial{y}} = {x}^{y - 1} + y .{x}^{y - 1} . \ln(x)$

$\implies \: \frac{ \partial}{ \partial{x}} \bigg(\frac{ \partial^{2} {u} }{ \partial{x} \partial{y}} \bigg) = (y - 1){x}^{y - 2} + y (y - 1).{x}^{y - 2} . \ln(x) + y. {x}^{y - 1}. \frac{1}{x}$

$\implies \: \frac{ \partial^{3} {u} }{ \partial{x} \partial{y} \partial{x}} = (y - 1){x}^{y - 2} + y (y - 1).{x}^{y - 2} . \ln(x) + y. {x}^{y - 2}$

$\implies \: \frac{ \partial^{3} {u} }{ \partial{x} \partial{y} \partial{x}} = \{(y - 1) + y (y - 1). \ln(x) + y \}.{x}^{y - 2}$

$\implies \: \frac{ \partial^{3} {u} }{ \partial{x} \partial{y} \partial{x}} = \{y - 1+ (y^{2} - y). \ln(x) + y \}.{x}^{y - 2}$

$\implies \: \frac{ \partial^{3} {u} }{ \partial{x} \partial{y} \partial{x}} = \{2y - 1+ (y^{2} - y). \ln(x) \}.{x}^{y - 2}$

Hence,

$\frac{ \partial^{3}{u} }{ \partial{ {x}^{2} } \partial{y}} = \frac{ \partial^{3}{u} }{ \partial{x } \partial{y} \partial{x}}$