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

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 \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\] to get the required answer.

Formula used:
\[\dfrac{\partial }{{\partial x}}\tan^{ - 1}\left( {x + y} \right) = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\dfrac{\partial }{{\partial x}}\left( {x + y} \right) = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\]
\[\dfrac{\partial }{{\partial y}}\tan^{ - 1}\left( {x + y} \right) = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\dfrac{\partial }{{\partial y}}\left( {x + y} \right) = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\]
\[\tan^{2}x + 1 = \sec^{2}x\]
\[\sin 2x = 2\sin x \cos x\]

Complete step by step solution:
The given inverse trigonometric equation is \[u = \tan^{ - 1}\left( {x + y} \right)\].
Let’s differentiate the above given equation partially with respect to the variable \[x\].
\[\dfrac{{\partial u}}{{\partial x}} = \dfrac{\partial }{{\partial x}}\tan^{ - 1}\left( {x + y} \right)\]
Apply the chain rule of differentiation.
\[\dfrac{{\partial u}}{{\partial x}} = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\dfrac{\partial }{{\partial x}}\left( {x + y} \right)\]
\[ \Rightarrow \dfrac{{\partial u}}{{\partial x}} = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\] \[.....\left( 1 \right)\]

Also, differentiate the above given equation partially with respect to the variable \[y\].
\[\dfrac{{\partial u}}{{\partial y}} = \dfrac{\partial }{{\partial y}}\tan^{ - 1}\left( {x + y} \right)\]
Apply the chain rule of differentiation.
\[\dfrac{{\partial u}}{{\partial y}} = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\dfrac{\partial }{{\partial y}}\left( {x + y} \right)\]
\[ \Rightarrow \dfrac{{\partial u}}{{\partial y}} = \dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}\] \[.....\left( 2 \right)\]

Now substitute the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] in the given expression \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\].
We get,
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = x\left( {\dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}} \right) + y\left( {\dfrac{1}{{1 + {{\left( {x + y} \right)}^2}}}} \right)\]
Simplify the above equation.
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \dfrac{{x + y}}{{1 + {{\left( {x + y} \right)}^2}}}\]
Resubstitute the value of \[\left( {x + y} \right)\] by using the given equation \[u = \tan^{ - 1}\left( {x + y} \right)\].
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\tan u}}{{1 + {{\left( {\tan u} \right)}^2}}}\]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\tan u}}{{1 + \tan^{2}u}}\]
Apply the trigonometric identity \[\tan^{2}x + 1 = \sec^{2}x\].
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\tan u}}{{\sec^{2}u}}\]
Use the trigonometric ratio \[\cos x = \dfrac{1}{{\sec x}}\] and \[\tan x = \dfrac{{\sin x}}{{\cos x}}\].
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \left( {\dfrac{{\sin u}}{{\cos u}} } \right)\cos^{2}u\]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \sin u \cos u\]
Now apply the trigonometric identity \[\sin 2x = 2\sin x \cos x\].
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \left( {\dfrac{1}{2}} \right)\sin 2u\]
Hence the correct option is B.

Note: Partial derivative is all most same as normal derivate. But in the partial derivative, we consider only one variable and other variables are treated as constant. When we find the partial derivative $\dfrac{\partial }{{\partial x}}$, then we consider \[x\] as a variable and \[y\] as a constant.