Question

If \[u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\], then find \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\].
A. \[ - \sin 2u\]
B. \[\sin 2u\]
C. \[\cos 2u\]
D. \[ - \cos 2u\]

Answer 1

Hint: First we rewrite the equation in the form \[\tan u = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\] and assume that \[z = \dfrac{{{x^3} + {y^3}}}{{x - y}}\]. Then we will check whether \[z\] is a homogeneous function. If it is a homogenous function then we apply Euler’s Theorem on \[z\]. Then find the partial derivative of \[\tan u = z\]. From these two equations, we can get \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\].

Formula used
Homogeneous function: If a homogeneous function can be written as \[{x^n}f\left( {\dfrac{y}{x}} \right)\].
Euler’s Theorem: Euler’s theorem is used to establish a relationship between the partial derivatives of a function and the product of the function with its degree.
\[\dfrac{d}{{dx}}\tan \theta = {\sec ^2}\theta \]
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\cos \theta = \dfrac{1}{{\sec \theta }}\]
\[2\sin \theta \cos \theta = \sin 2\theta \]

Complete step by step solution
Given equation is
\[u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
Apply the formula \[\alpha = {\tan ^{ - 1}}\beta \Rightarrow \tan \alpha = \beta \]
\[\tan u = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
Let \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
Taking common \[{x^3}\] from the numerator and \[x\] from the denominator
\[ \Rightarrow z = \dfrac{{{x^3}\left( {1 + {{\left( {\dfrac{y}{x}} \right)}^3}} \right)}}{{x\left( {1 - \dfrac{y}{x}} \right)}}\]
\[ \Rightarrow z = {x^2}f\left( {\dfrac{y}{x}} \right)\]
The degree of \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\] is 2.
Now applying Euler’s Theorem in \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
\[x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = 2z\] ……(1) [Since 2 is degree of \[z\]]
Now putting \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\] in \[\tan u = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
\[\tan u = z\]
Apply partial differentiate with respect to \[x\] on \[\tan u = z\]
\[{\sec ^2}u\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial z}}{{\partial x}}\]
Apply partial differentiate with respect to \[y\] on \[\tan u = z\]
\[{\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\partial z}}{{\partial y}}\]
Now putting the value of \[\dfrac{{\partial z}}{{\partial x}}\] and \[\dfrac{{\partial z}}{{\partial y}}\] in the equation \[x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = 2z\].
\[x{\sec ^2}u\dfrac{{\partial u}}{{\partial x}} + y{\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = 2z\]
Now we will put \[\tan u = z\]
\[x{\sec ^2}u\dfrac{{\partial u}}{{\partial x}} + y{\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = 2\tan u\]
Divide both sides by \[{\sec ^2}u\]
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\dfrac{{\tan u}}{{{{\sec }^2}u}}\]
Apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cos \theta = \dfrac{1}{{\sec \theta }}\]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2 \cdot \dfrac{{\sin u}}{{\cos u}} \cdot {\cos ^2}u\]
Cancel out \[\cos u\]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\sin u\cos u\]
Now applying \[2\sin \theta \cos \theta = \sin 2\theta \]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \sin 2u\]
Hence option B is the correct option.

Note: Sometimes students do mistake to calculate the degree of \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]. They take the degree as 3 because the highest power of the variable is 3. A homogenous function can be written as \[{x^n}f\left( {\dfrac{y}{x}} \right)\] and the degree of the function is \[n\]. \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\] can be written as \[z = {x^2}f\left( {\dfrac{y}{x}} \right)\]. Thus, the degree of \[z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\] is 2.