Question

if z(x+y) =x²+y² show that (dz/dx - dz/dy) ² = 4(1-dz/dx-dz/dy)​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{z=\dfrac{x^2+y^2}{x+y}}$

$\underline{\textbf{To prove:}}$

$\mathsf{\left(\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)^2=4\left(1-\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Concept used:}}$

$\boxed{\mathsf{\dfrac{d\left(\dfrac{u}{v}\right)}{dx}=\dfrac{v\;\dfrac{du}{dx}-u\;\dfrac{dv}{dx}}{v^2}}}$

$\mathsf{Consider,}$

$\mathsf{z=\dfrac{x^2+y^2}{x+y}}$

$\textsf{Differentiate partially with respect to 'x'}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}=\dfrac{(x+y)2x-(x^2+y^2).1}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}=\dfrac{2x^2+2xy-x^2-y^2}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}=\dfrac{x^2+2xy-y^2}{(x+y)^2}}$

$\mathsf{z=\dfrac{x^2+y^2}{x+y}}$

$\textsf{Differentiate partially with respect to 'y'}$

$\mathsf{\dfrac{\partial\,z}{\partial\,y}=\dfrac{(x+y)2y-(x^2+y^2).1}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,y}=\dfrac{2xy+2y^2-x^2-y^2}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,y}=\dfrac{-x^2+2xy+y^2}{(x+y)^2}}$

$\mathsf{Now,}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{x^2+2xy-y^2}{(x+y)^2}-\left(\dfrac{-x^2+2xy+y^2}{(x+y)^2}\right)}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{x^2+2xy-y^2+x^2-2xy-y^2}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{2x^2-2y^2}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{2(x^2-y^2)}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{2(x+y)(x-y)}{(x+y)^2}}$

$\mathsf{\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}=\dfrac{2(x-y)}{x+y}}$

$\mathsf{\left(\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)^2=\dfrac{4(x-y)^2}{(x+y)^2}}$---------(1)

$\mathsf{4\left(1-\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)}$

$\mathsf{=4\left(1-\left(\dfrac{x^2+2xy-y^2}{(x+y)^2}\right)-\left(\dfrac{-x^2+2xy+y^2}{(x+y)^2}\right)\right)}$

$\mathsf{=4\left(1-\dfrac{4xy}{(x+y)^2}\right)}$

$\mathsf{=4\left(\dfrac{(x+y)^2-4xy}{(x+y)^2}\right)}$

$\mathsf{=4\left(\dfrac{(x-y)^2}{(x+y)^2}\right)}$---------------(2)

$\textsf{From (1) and (2), we get}$

$\mathsf{\left(\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)^2=4\left(1-\dfrac{\partial\,z}{\partial\,x}-\dfrac{\partial\,z}{\partial\,y}\right)}$

Answer 2

Given:

z(x + y) =x²+y²

we can write it as

$z=\frac{x^{2}+y^{2}}{x+y}$

Now solve this

$z=\frac{x^{2}+y^{2}}{x+y}$

$\text { Differentiate partially with respect to ' } x \text { ' }$

$\begin{array}{l}\frac{\partial z}{\partial x}=\frac{(x+y) 2 x-\left(x^{2}+y^{2}\right) .1}{(x+y)^{2}} \\\end{array}$

$\frac{\partial z}{\partial x}=\frac{2 x^{2}+2 x y-x^{2}-y^{2}}{(x+y)^{2}}$

$\begin{array}{l}\frac{\partial z}{\partial x}=\frac{x^{2}+2 x y-y^{2}}{(x+y)^{2}} \\\end{array}$

$z=\frac{x^{2}+y^{2}}{x+y}$

$\text { Differentiate partially with respect to ' } y \text { ' }$

$\begin{array}{l}\frac{\partial z}{\partial y}=\frac{(x+y) 2 y-\left(x^{2}+y^{2}\right) .1}{(x+y)^{2}} \\\end{array}$

$\frac{\partial z}{\partial y}=\frac{2 x y+2 y^{2}-x^{2}-y^{2}}{(x+y)^{2}}$

$\frac{\partial z}{\partial y}=\frac{-x^{2}+2 x y+y^{2}}{(x+y)^{2}}$

$\begin{array}{l}\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{x^{2}+2 x y-y^{2}}{(x+y)^{2}}-\left(\frac{-x^{2}+2 x y+y^{2}}{(x+y)^{2}}\right) \\\end{array}$

$\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{x^{2}+2 x y-y^{2}+x^{2}-2 x y-y^{2}}{(x+y)^{2}}$

$\begin{array}{l}\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{2 x^{2}-2 y^{2}}{(x+y)^{2}} \\\end{array}$

$\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{2\left(x^{2}-y^{2}\right)}{(x+y)^{2}}$

$\begin{array}{l}\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{2(x+y)(x-y)}{(x+y)^{2}} \\\end{array}$

$\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{2(x-y)}{x+y}$

$\left(\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)^{2}=\frac{4(x-y)^{2}}{(x+y)^{2}}$      -(1)

$4\left(1-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)=4\left(1-\left(\frac{x^{2}+2 x y-y^{2}}{(x+y)^{2}}\right)-\left(\frac{-x^{2}+2 x y+y^{2}}{(x+y)^{2}}\right)\right)$

$\begin{array}{l}=4\left(1-\frac{4 x y}{(x+y)^{2}}\right) \\\end{array}$

$=4\left(\frac{(x+y)^{2}-4 x y}{(x+y)^{2}}\right)$

$=4\left(\frac{(x-y)^{2}}{(x+y)^{2}}\right)$     -(2)

From (1) and (2) we get

$\left(\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)^{2}=4\left(1-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)$

Hence its prove