Question

If $z = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$, then check $\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = 1$.
A. True
B. False

Answer 1

Hint: First we will first order partial derivative of $z = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ with respect to $x$ and $y$. Then again find the partial derivative of the first-order derivative with respect to $x$ and $y$. Then add the partial derivative to verify the given equation $\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = 1$.

Formula Used:
$\dfrac{\partial }{{\partial x}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$
$\dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$

Complete step by step solution:
Given that, $z = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
Now find the partial derivative of $z = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ with respect to $x$.
$\dfrac{{\partial z}}{{\partial x}} = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( { - \dfrac{y}{{{x^2}}}} \right)$
$ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = \dfrac{{{x^2}}}{{{x^2} + {y^2}}}\left( { - \dfrac{y}{{{x^2}}}} \right)$
$ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = - \dfrac{y}{{{x^2} + {y^2}}}$
Again, find the partial derivative with respect to $x$.
$\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = \dfrac{y}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}\left( {2x} \right)$
$ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = \dfrac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$ …..(i)
Again, find the partial derivative of $z = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ with respect to $y$.
$\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{1 + {{\left( {\dfrac{y}{x}} \right)}^2}}}\left( {\dfrac{1}{x}} \right)$
$ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{{{x^2}}}{{{x^2} + {y^2}}} \cdot \left( {\dfrac{1}{x}} \right)$
$ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{x}{{{x^2} + {y^2}}}$
Again, find the partial derivative with respect to $y$.
$ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = - \dfrac{x}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}\left( {2y} \right)$
$ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = - \dfrac{{2yx}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$ ….(ii)
Add equation (i) and equation (ii)
$\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = \dfrac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} + \left( { - \dfrac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right)$
$ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$
So the given statement is false.

Note: Partial derivative is almost the same as normal derivative. But in the partial derivative we are considering 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.