Question

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

Answer 1

Hint: First we will find reverse of inverse trigonometry. Then find the partial derivative of the equation with respect to \[x\] and \[y\]. Then calculate the value of \[x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}}\].

Formula used:
\[\dfrac{\partial }{{\partial x}}\left( {\sin x} \right) = \cos x\]
\[\dfrac{\partial }{{\partial x}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{\partial u}}{{\partial x}} - u\dfrac{{\partial v}}{{\partial x}}}}{{{v^2}}}\]

Complete step by step solution:
Given equation is \[z = {\sin ^{ - 1}}\left( {\dfrac{{x + y}}{{\sqrt x + \sqrt y }}} \right)\].
Apply the reverse formula of trigonometry inverse
\[\sin z = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}\] …….(i)
Calculate the partial derivative of equation (i) with respect to \[x\]
\[\cos z\dfrac{{\partial z}}{{\partial x}} = \dfrac{\partial }{{\partial x}}\left( {\dfrac{{x + y}}{{\sqrt x + \sqrt y }}} \right)\]
Apply the quotient formula on the right side:
\[ \Rightarrow \cos z\dfrac{{\partial z}}{{\partial x}} = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\dfrac{\partial }{{\partial x}}\left( {x + y} \right) - \left( {x + y} \right)\dfrac{\partial }{{\partial x}}\left( {\sqrt x + \sqrt y } \right)}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \cos z\dfrac{{\partial z}}{{\partial x}} = \dfrac{{\left( {\sqrt x + \sqrt y } \right) \cdot 1 - \left( {x + y} \right) \cdot \dfrac{1}{{2\sqrt x }}}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
Divide both sides by \[\cos z\].
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{\dfrac{{2\sqrt x \left( {\sqrt x + \sqrt y } \right) - \left( {x + y} \right)}}{{2\sqrt x }}}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{2\sqrt x \left( {\sqrt x + \sqrt y } \right) - \left( {x + y} \right)}}{{2\sqrt x {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{2x + 2\sqrt {xy} - x - y}}{{2\sqrt x {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{x + 2\sqrt {xy} - y}}{{2\sqrt x {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\] ……..(ii)
Calculate the partial derivative of equation (i) with respect to \[y\]
\[\cos z\dfrac{{\partial z}}{{\partial y}} = \dfrac{\partial }{{\partial y}}\left( {\dfrac{{x + y}}{{\sqrt x + \sqrt y }}} \right)\]
Apply the quotient formula on the right side:
\[ \Rightarrow \cos z\dfrac{{\partial z}}{{\partial y}} = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\dfrac{\partial }{{\partial y}}\left( {x + y} \right) - \left( {x + y} \right)\dfrac{\partial }{{\partial y}}\left( {\sqrt x + \sqrt y } \right)}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \cos z\dfrac{{\partial z}}{{\partial x}} = \dfrac{{\left( {\sqrt x + \sqrt y } \right) \cdot 1 - \left( {x + y} \right) \cdot \dfrac{1}{{2\sqrt y }}}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
Divide both sides by \[\cos z\].
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{\dfrac{{2\sqrt y \left( {\sqrt x + \sqrt y } \right) - \left( {x + y} \right)}}{{2\sqrt y }}}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{2\sqrt y \left( {\sqrt x + \sqrt y } \right) - \left( {x + y} \right)}}{{2\sqrt y {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{2\sqrt {xy} + 2y - x - y}}{{2\sqrt y {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{y + 2\sqrt {xy} - x}}{{2\sqrt y {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]…..(iii)
Multiply \[x\] with equation (ii) and multiply \[y\] with equation (iii) and add them
\[x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{\cos z}} \cdot \dfrac{{x\left( {x + 2\sqrt {xy} - y} \right)}}{{2\sqrt x {{\left( {\sqrt x + \sqrt y } \right)}^2}}} + \dfrac{1}{{\cos z}} \cdot \dfrac{{y\left( {y + 2\sqrt {xy} - x} \right)}}{{2\sqrt y {{\left( {\sqrt x + \sqrt y } \right)}^2}}}\]
Take common \[\dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\] from the right side expression
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left[ {\dfrac{{\sqrt x \cdot \sqrt x \left( {x + 2\sqrt {xy} - y} \right)}}{{\sqrt x }} + \dfrac{{\sqrt y \cdot \sqrt y \left( {y + 2\sqrt {xy} - x} \right)}}{{\sqrt y }}} \right]\]
Cancel out same terms from the numerator and denominator
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left[ {\sqrt x \left( {x + 2\sqrt {xy} - y} \right) + \sqrt y \left( {y + 2\sqrt {xy} - x} \right)} \right]\]
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left[ {x\sqrt x + 2x\sqrt y - y\sqrt x + y\sqrt y + 2y\sqrt x - x\sqrt y } \right]\]
Subtract like on the numerator
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left[ {x\sqrt x + x\sqrt y + y\sqrt y + y\sqrt x } \right]\]
Apply factor method
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left[ {x\left( {\sqrt x + \sqrt y } \right) + y\left( {\sqrt y + \sqrt x } \right)} \right]\]
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\left( {\sqrt x + \sqrt y } \right)\left( {x + y} \right)\]
Cancel out \[\left( {\sqrt x + \sqrt y } \right)\] from the denominator and numerator from the right-side expression
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z}} \cdot \dfrac{{\left( {x + y} \right)}}{{\left( {\sqrt x + \sqrt y } \right)}}\]
From equation (i) we get \[\sin z = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}\]
Substitute \[\sin z = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}\] in the above equation
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{{2\cos z}} \cdot \sin z\]
We know that \[\dfrac{{\sin x}}{{\cos x}} = \tan x\].
\[ \Rightarrow x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = \dfrac{1}{2}\tan z\]
Hence option B is the correct option.

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 variable and \[y\] as a constant.