Question

Найти все частные производные производные 2 порядка u=xyln(y/z)​

Answer 1

$u=xy\ln\dfrac{y}{z}$

Сначала найдем частные производные первого порядка:

$u'_x=\left(xy\ln\dfrac{y}{z}\right)'_x=\boxed{y\ln\dfrac{y}{z}}$

$u'_y=\left(xy\ln\dfrac{y}{z}\right)'_y=x\left(y\ln\dfrac{y}{z}\right)'_y=x\left(y'_y\cdot\ln\dfrac{y}{z}+y\cdot\left(\ln\dfrac{y}{z}\right)'_y\right)=$

$=x\left(1\cdot\ln\dfrac{y}{z}+y\cdot\dfrac{z}{y} \cdot\left(\dfrac{y}{z}\right)'_y\right)=x\left(\ln\dfrac{y}{z}+z\cdot\dfrac{1}{z}\right)=\boxed{x\left(\ln\dfrac{y}{z}+1\right)}$

$u'_z=\left(xy\ln\dfrac{y}{z}\right)'_z=xy\left(\ln\dfrac{y}{z}\right)'_z=xy\cdot \dfrac{z}{y}\cdot\left(\dfrac{y}{z}\right)'_z=xz\cdot y\cdot\left(\dfrac{1}{z}\right)'_z=$

$=xyz\cdot \left(-\dfrac{1}{z^2}\right)=\boxed{-\dfrac{xy}{z}}$

Теперь найдем частные производные второго порядка:

$u''_{xx}=(u'_x)'_x=\left(y\ln\dfrac{y}{z}\right)'_x=\boxed{0}$

$u''_{yy}=(u'_y)'_y=\left(x\left(\ln\dfrac{y}{z}+1\right)\right)'_y=x\left(\ln\dfrac{y}{z}+1\right)'_y=x\left(\dfrac{z}{y}\cdot\left(\dfrac{y}{z}\right)'_y+0\right)=$

$=\dfrac{xz}{y}\cdot\dfrac{1}{z}=\boxed{\dfrac{x}{y}}$

$u''_{zz}=(u'_z)'_z=\left(-\dfrac{xy}{z}\right)'_z=-xy\cdot\left(\dfrac{1}{z}\right)'_z=-xy\cdot\left(-\dfrac{1}{z^2}\right)=\boxed{\dfrac{xy}{z^2}}$

$u''_{xy}=u''_{yx}=(u'_y)'_x=\left(x\left(\ln\dfrac{y}{z}+1\right)\right)'_x=\boxed{\ln\dfrac{y}{z}+1}$

$u''_{xz}=u''_{zx}=(u'_z)'_x=\left(-\dfrac{xy}{z}\right)'_x=\boxed{-\dfrac{y}{z}}$

$u''_{yz}=u''_{zy}=(u'_z)'_y=\left(-\dfrac{xy}{z}\right)'_y=\boxed{-\dfrac{x}{z}}$

Answer 2

Finding the Partial Derivatives of the Second Order

To find the partial derivatives of the second order for the function u = xyln(y/z), we need to calculate the partial derivatives with respect to each variable twice.

Let's start by finding the first-order partial derivatives:

- Partial derivative with respect to x (u_x): To find u_x, we differentiate u with respect to x while treating y and z as constants. Taking the derivative of xyln(y/z) with respect to x, we get: u_x = yln(y/z).

- Partial derivative with respect to y (u_y): To find u_y, we differentiate u with respect to y while treating x and z as constants. Taking the derivative of xyln(y/z) with respect to y, we get: u_y = x(ln(y/z) + 1).

- Partial derivative with respect to z (u_z): To find u_z, we differentiate u with respect to z while treating x and y as constants. Taking the derivative of xyln(y/z) with respect to z, we get: u_z = -xy/(z).

Now, let's find the second-order partial derivatives:

- Second-order partial derivative with respect to x (u_xx): To find u_xx, we differentiate u_x with respect to x while treating y and z as constants. Taking the derivative of yln(y/z) with respect to x, we get: u_xx = 0.

- Second-order partial derivative with respect to y (u_yy): To find u_yy, we differentiate u_y with respect to y while treating x and z as constants. Taking the derivative of x(ln(y/z) + 1) with respect to y, we get: u_yy = 0.

- Second-order partial derivative with respect to z (u_zz): To find u_zz, we differentiate u_z with respect to z while treating x and y as constants. Taking the derivative of -xy/(z) with respect to z, we get: u_zz = 2xy/(z^3).

- Second-order partial derivative with respect to x and y (u_xy): To find u_xy, we differentiate u_x with respect to y while treating x and z as constants. Taking the derivative of yln(y/z) with respect to y, we get: u_xy = ln(y/z).

- Second-order partial derivative with respect to x and z (u_xz): To find u_xz, we differentiate u_x with respect to z while treating x and y as constants. Taking the derivative of yln(y/z) with respect to z, we get: u_xz = -y/(z).

- Second-order partial derivative with respect to y and z (u_yz): To find u_yz, we differentiate u_y with respect to z while treating x and y as constants. Taking the derivative of x(ln(y/z) + 1) with respect to z, we get: u_yz = -x/(z).

Now we have found all the second-order partial derivatives for the function u = xyln(y/z).

Please let me know if there's anything else I can help you with!