If \[u = xy{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right)\] , then what is the value of \[x{u_x} + y{u_y}\]?
A. \[u\]
B. \[2u\]
C. \[3u\]
D. \[\dfrac{u}{3}\]

Hint: First, differentiate the given equation partially with respect to \[x\]. Again, differentiate the given equation partially with respect to \[y\]. Substitute both differential equations in the given expression \[x{u_x} + y{u_y}\] to get the required answer.

Formula used:
\[\dfrac{\partial }{{\partial x}}\tan^{ - 1}\left( {\dfrac{y}{x}} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\dfrac{\partial }{{\partial x}}\left( {\dfrac{y}{x}} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( { - \dfrac{y}{{{x^2}}}} \right)\]
\[\dfrac{\partial }{{\partial y}}\tan^{ - 1}\left( {\dfrac{y}{x}} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\dfrac{\partial }{{\partial x}}\left( {\dfrac{y}{x}} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( {\dfrac{1}{x}} \right)\]
Product rule of differentiation: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given inverse trigonometric equation is \[u = xy{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right)\].
Let’s differentiate the above equation partially.
Consider \[x\] as one term and \[y{}^2ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)\] as other term.
Differentiate the above given equation partially with respect to the variable \[x\].
Apply the product rule of differentiation.
\[{u_x} = \left( {\dfrac{\partial }{{\partial x}}x} \right)y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + x\left( {\dfrac{\partial }{{\partial x}}y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right)} \right)\]
Apply the chain rule for the differentiation of the second term.
\[{u_x} = y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + x{y^2}\left( {\dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( { - \dfrac{y}{{{x^2}}}} \right)} \right)\]
\[ \Rightarrow {u_x} = y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + x{y^2}\left( {\dfrac{{{x^2}}}{{{x^2} + {y^2}}}\left( { - \dfrac{y}{{{x^2}}}} \right)} \right)\]
\[ \Rightarrow {u_x} = y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) - \dfrac{{x{y^3}}}{{\left( {{x^2} + {y^2}} \right)}}\] \[.....\left( 1 \right)\]

Now differentiate the above given equation partially with respect to the variable \[y\].
Consider \[y{}^2\] as one term and \[x\tan^{ - 1}\left( {\dfrac{y}{x}} \right)\] as other term.
Apply the product rule of differentiation.
\[{u_y} = \left({\dfrac{\partial }{{\partial y}}y{}^2} \right)x\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + {y^2}\left( {\dfrac{\partial }{{\partial y}}x\tan^{ - 1}\left( {\dfrac{y}{x}} \right)} \right)\]
Apply the chain rule for the differentiation of the second term.
\[{u_y} = 2xy\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + x{y^2}\left( {\dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( {\dfrac{1}{x}} \right)} \right)\]
\[ \Rightarrow {u_y} = 2xy\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + x{y^2}\left( {\dfrac{{{x^2}}}{{{x^2} + {y^2}}}\left( {\dfrac{1}{x}} \right)} \right)\]
\[ \Rightarrow {u_y} = 2xy\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + \dfrac{{{x^2}{y^2}}}{{\left( {{x^2} + {y^2}} \right)}}\] \[.....\left( 2 \right)\]

Now substitute the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] in the given expression \[x{u_x} + y{u_y}\].
We get,
\[x{u_x} + y{u_y} = x\left[ {y{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) - \dfrac{{x{y^3}}}{{\left( {{x^2} + {y^2}} \right)}}} \right] + y\left[ {2xy\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + \dfrac{{{x^2}{y^2}}}{{\left( {{x^2} + {y^2}} \right)}}} \right]\]
Simplify the above equation.
\[x{u_x} + y{u_y} = xy{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) - \dfrac{{{x^2}{y^3}}}{{\left( {{x^2} + {y^2}} \right)}} + 2x{y^2}\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + \dfrac{{{x^2}{y^3}}}{{\left( {{x^2} + {y^2}} \right)}}\]
\[ \Rightarrow x{u_x} + y{u_y} = xy{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right) + 2x{y^2}\tan^{ - 1}\left( {\dfrac{y}{x}} \right)\]
\[ \Rightarrow x{u_x} + y{u_y} = 3x{y^2}\tan^{ - 1}\left( {\dfrac{y}{x}} \right)\]
\[ \Rightarrow x{u_x} + y{u_y} = 3u\] \[\left[ {\because u = xy{}^2\tan^{ - 1}\left( {\dfrac{y}{x}} \right)} \right]\]
Hence the correct option is C.

Note: Students often get confused and directly apply the product rule without using the chain rule of differentiation.
The chain rule is used to take derivatives of composites of functions and this happens by chaining together their derivatives.
Chain rule: If \[F\left( x \right) = f\left( {g\left( x \right)} \right)\], then \[F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\]