Respuesta :
Answer:
The value of  [tex]\frac{\partial z}{\partial x}[/tex]  at (1,1,1) is 1.
Step-by-step explanation:
Given,
[tex] z''x+8xy-2yz-7=0\hfill (1)[/tex]
wnere, z=f(x,y).
We have to find,
[tex]\frac{\partial z}{\partial x}[/tex]
From (1) letting z'=m we get,
[tex]m^2-2ym+(8xy-7)=0[/tex]
[tex]\implies m=\frac{-(-2y)\pm \sqrt{(-2y)^2-4\times(8xy-7)}}{2}[/tex]
[tex]=\frac{2y\pm \sqrt{4y^2-8xy+7}}{2}[/tex]
[tex]\therefore m=z'=\frac{dz}{dx}=y\pm \sqrt{y^2=8xy+7}[/tex]
[tex]dz=y dx\pm \sqrt{y^2-8xy+7}dx[/tex]
Integrate by letting [tex]u=y^2-8xy+7[/tex] we get,
[tex]\int dz=xy\pm \int \frac{\sqrt{u}}{8y}[/tex]
[tex]z(x,y)=xy\pm \frac{1}{8y}\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}[/tex]
[tex]=xy\pm\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}[/tex]
Now differentiate partially with respect to x we get,
[tex]\frac{\partial z}{\partial x}=y\pm\frac{\partial}{\partial x}\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}[/tex]
[tex]=y\pm\frac{1}{12y}\frac{\partial}{\partial x}(y^2-8xy+7)^{\frac{5}{2}[/tex]
Letting [tex]u=y^2-8xy+7, f=u^\frac{5}{2}[/tex] and applying chain rule,
[tex]\frac{df(u)}{dx}=\frac{df}{du}\frac{du}{dx}[/tex] we get,
[tex]\frac{\partial z}{\partial x}=y\pm\frac{1}{12y}\frac{\partial}{\partial u}u^{\frac{5}{2}}\frac{\partial}{\partial x}(y^2-8xy+7)[/tex]
[tex]=y\pm\frac{5}{3}(y^2-8xy+7)^{\frac{3}{2}[/tex]
At (1,1,1),
[tex]\Big[\frac{\partial z}{\partial x}\Big]_{(1,1,1)}=1[/tex]
Hence the result.
