Question

If $u = {x^2}{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) - {y^2}{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)$ then find $\dfrac{{{\partial ^2}u}}{{\partial x\partial y}}$.

Answer 1

Hint: The required derivative is a mixed derivative. First find the partial derivative of $u$ with respect to $y$ keeping $x$ as constant. You will get the expression for $\dfrac{{\partial u}}{{\partial y}}$. After that find the partial derivative of this expression with respect to $x$ keeping $y$ as constant to obtain the expression for $\dfrac{{{\partial ^2}u}}{{\partial x\partial y}}$.

Formula Used:
If $u = f\left( {x,y} \right) \cdot g\left( {x,y} \right)$, then
$\dfrac{{\partial u}}{{\partial x}} = f\left( {x,y} \right)\dfrac{\partial }{{\partial x}}\left\{ {g\left( {x,y} \right)} \right\} + g\left( {x,y} \right)\dfrac{\partial }{{\partial x}}\left\{ {f\left( {x,y} \right)} \right\}$ and $\dfrac{{\partial u}}{{\partial y}} = f\left( {x,y} \right)\dfrac{\partial }{{\partial y}}\left\{ {g\left( {x,y} \right)} \right\} + g\left( {x,y} \right)\dfrac{\partial }{{\partial y}}\left\{ {f\left( {x,y} \right)} \right\}$
If$z = f\left( {u,v} \right)$, where $u = g\left( {x,y} \right)$ and $v = h\left( {x,y} \right)$, then
$\dfrac{{\partial z}}{{\partial x}} = \dfrac{{\partial z}}{{\partial u}} \cdot \dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial z}}{{\partial v}} \cdot \dfrac{{\partial v}}{{\partial x}}$ and $\dfrac{{\partial z}}{{\partial y}} = \dfrac{{\partial z}}{{\partial u}} \cdot \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial z}}{{\partial v}} \cdot \dfrac{{\partial v}}{{\partial y}}$
If $y = {\tan ^{ - 1}}x$, then $\dfrac{{\partial y}}{{\partial x}} = \dfrac{1}{{1 + {x^2}}}$
If $y = {x^n}$, then $\dfrac{{\partial y}}{{\partial x}} = n{x^{n - 1}}$

Complete step by step solution:
Given that $u = {x^2}{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) - {y^2}{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)$
Differentiating partially with respect to $y$, we get
$\dfrac{{\partial u}}{{\partial y}} = \dfrac{\partial }{{\partial y}}\left\{ {{x^2}{{\tan }^{ - 1}}\left( {\dfrac{y}{x}} \right) - {y^2}{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}$
       $ \dfrac{\partial }{{\partial y}}\left\{ {{x^2}{{\tan }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right\} - \dfrac{\partial }{{\partial y}}\left\{ {{y^2}{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}\\ \Rightarrow {x^2}\dfrac{\partial }{{\partial y}}\left\{ {{{\tan }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right\} - \left[ {{y^2} \times \dfrac{\partial }{{\partial y}}\left\{ {{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\} + {{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right) \times \dfrac{\partial }{{\partial y}}\left\{ {{y^2}} \right\}} \right]\\ \Rightarrow {x^2} \times \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\dfrac{\partial }{{\partial y}}\left( {\dfrac{y}{x}} \right) - \left[ {{y^2} \times \dfrac{1}{{1 + {{\left( {\dfrac{x}{y}} \right)}^2}}} \times \dfrac{\partial }{{\partial y}}\left( {\dfrac{x}{y}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right) \times 2y} \right]\\ \Rightarrow {x^{{2}}} \times \dfrac{{{x^2}}}{{{x^2} + {y^2}}} \times \dfrac{1}{{{x}}} \times \dfrac{\partial }{{\partial y}}\left( y \right) - \left[ {{y^2} \times \dfrac{{{y^2}}}{{{y^2} + {x^2}}} \times x \times \dfrac{\partial }{{\partial y}}\left( {\dfrac{1}{y}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right) \times 2y} \right]\\ \Rightarrow \dfrac{{{x^3}}}{{{x^2} + {y^2}}} \times 1 - \left[ {\dfrac{{x{y^4}}}{{{x^2} + {y^2}}} \times \left( { - \dfrac{1}{{{y^2}}}} \right) + 2y{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right]\\ \Rightarrow \dfrac{{{x^3}}}{{{x^2} + {y^2}}} + \dfrac{{x{y^2}}}{{{x^2} + {y^2}}} - 2y{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)\\ \Rightarrow \dfrac{{{x^3} + x{y^2}}}{{{x^2} + {y^2}}} - 2y{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)\\ \Rightarrow \dfrac{{x\left( {{x^2} + {y^2}} \right)}}{{{x^2} + {y^2}}} - 2y{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)\\ \Rightarrow x - 2y{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)$
Differentiating the above equation with respect to $x$, we get
$\dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{\partial }{{\partial x}}\left\{ {x - 2y{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}$
$ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial x\partial y}} = \dfrac{\partial }{{\partial x}}\left( x \right) - 2\dfrac{\partial }{{\partial x}}\left\{ {y{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}$
$ 1 - 2\left[ {y \times \dfrac{\partial }{{\partial x}}\left\{ {{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}} \right]\\ \Rightarrow 1 - 2y\dfrac{\partial }{{\partial x}}\left\{ {{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}\\ \Rightarrow 1 - 2y \times \dfrac{1}{{1 + {{\left( {\dfrac{x}{y}} \right)}^2}}} \times \dfrac{\partial }{{\partial x}}\left( {\dfrac{x}{y}} \right)\\ \Rightarrow 1 - 2y \times \dfrac{{{y^2}}}{{{y^2} + {x^2}}} \times \dfrac{1}{y} \times \dfrac{\partial }{{\partial x}}\left( x \right)\\ \Rightarrow 1 - \dfrac{{2{y^2}}}{{{x^2} + {y^{^2}}}} \times 1\\ \Rightarrow 1 - \dfrac{{2{y^2}}}{{{x^2} + {y^{^2}}}}\\ \Rightarrow \dfrac{{{x^2} + {y^{^2}} - 2{y^{^2}}}}{{{x^2} + {y^{^2}}}}\\ \Rightarrow \dfrac{{{x^2} - {y^{^2}}}}{{{x^2} + {y^{^2}}}}$

Hence, $\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} = \dfrac{{{x^2} - {y^{^2}}}}{{{x^2} + {y^{^2}}}}$

Note: Don’t mistake to use the formula properly and it is a huge calculation-based problem. So, stay calm and keep focus while solving this type of problems. If you differentiate $u$ partially with respect to $y$ keeping $x$ as constant, you will get the expression for $\dfrac{{\partial u}}{{\partial y}}$ and If you differentiate $u$ partially with respect to $x$ keeping $y$ as constant, you will get the expression for $\dfrac{{\partial u}}{{\partial x}}$. To get the expression for $\dfrac{{{\partial ^2}u}}{{\partial x\partial y}}$, you have to differentiate the expression for $\dfrac{{\partial u}}{{\partial y}}$ with respect to $x$. Many students differentiate the expression for $\dfrac{{\partial u}}{{\partial x}}$ and then differentiate it with respect to $y$ and obtain the wrong answer.