Question

Find the value ofStartFraction partial derivative z Over partial derivative x EndFraction∂z∂xFind the value of at the point (1,1,1) if the equation 8xy+z"x- 2yz -7 defines z as a function of the two independent variables x and y and the partial derivative exists The value of dx at point (1,1,1) is□ Type an integer or a simplified fraction.)

Answer 1

Answer:

The value of   $\frac{\partial z}{\partial x}$  at (1,1,1) is 1.

Step-by-step explanation:

Given,

$z''x+8xy-2yz-7=0\hfill (1)$

wnere, z=f(x,y).

We have to find,

$\frac{\partial z}{\partial x}$

From (1) letting z'=m we get,

$m^2-2ym+(8xy-7)=0$

$\implies m=\frac{-(-2y)\pm \sqrt{(-2y)^2-4\times(8xy-7)}}{2}$

$=\frac{2y\pm \sqrt{4y^2-8xy+7}}{2}$

$\therefore m=z'=\frac{dz}{dx}=y\pm \sqrt{y^2=8xy+7}$

$dz=y dx\pm \sqrt{y^2-8xy+7}dx$

Integrate by letting $u=y^2-8xy+7$ we get,

$\int dz=xy\pm \int \frac{\sqrt{u}}{8y}$

$z(x,y)=xy\pm \frac{1}{8y}\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}$

$=xy\pm\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}$

Now differentiate partially with respect to x we get,

$\frac{\partial z}{\partial x}=y\pm\frac{\partial}{\partial x}\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}$

$=y\pm\frac{1}{12y}\frac{\partial}{\partial x}(y^2-8xy+7)^{\frac{5}{2}$

Letting $u=y^2-8xy+7, f=u^\frac{5}{2}$ and applying chain rule,

$\frac{df(u)}{dx}=\frac{df}{du}\frac{du}{dx}$ we get,

$\frac{\partial z}{\partial x}=y\pm\frac{1}{12y}\frac{\partial}{\partial u}u^{\frac{5}{2}}\frac{\partial}{\partial x}(y^2-8xy+7)$

$=y\pm\frac{5}{3}(y^2-8xy+7)^{\frac{3}{2}$

At (1,1,1),

$\Big[\frac{\partial z}{\partial x}\Big]_{(1,1,1)}=1$

Hence the result.