 William Cleghorn

2022-01-17

Compute the first-order partial derivatives.
$z=\frac{x}{y}$ Toni Scott

Expert

Step 1
When calculating partial derivatives for a function with several variables, one variable is assumed to be constant while the other variable is differentiated. For instance, using y as a constant and differentiating with regard to the other variable, a function with two variables, x, and y, can have its partial derivative with respect to x determined.
By treating x as a constant and differentiating with respect to y, one can calculate the partial derivative with respect to y. The function for the described situation consists of two variables. So either partial derivative with respect to x or partial derivative with respect to x will be the first-order partial derivatives.
Step 2
Given function is $z=\frac{x}{y}$. First calculate the partial derivative with respect to x. Here, treat y as a constant.
${z}_{x}=\frac{\partial }{\partial x}\left(\frac{x}{y}\right)$
$=\frac{1}{y}\frac{\partial }{\partial x}\left(x\right)$
$=\frac{1}{y}\cdot 1$
$=\frac{1}{y}$
Do a partial derivative calculation with respect to y now. Here, consider x to be a constant.
${z}_{y}=\frac{\partial }{\partial y}\left(\frac{x}{y}\right)$
$=x\frac{\partial }{\partial y}\left(\frac{1}{y}\right)$
$=x\cdot \frac{\partial }{\partial y}\left({y}^{-1}\right)$
$=x\cdot \left(-{y}^{-2}\right)$
$=-\frac{x}{{y}^{2}}$ servidopolisxv

Expert

The required is to compute the first-order partial derivatives:
$z=\frac{x}{y}$
The first-order partial derivatives:
$\therefore {z}_{x}=\frac{\partial }{\partial x}\left(\frac{x}{y}\right)$
$=\frac{1}{y}$
$\therefore {z}_{y}=\frac{\partial }{\partial y}\left(\frac{x}{y}\right)$
$=\frac{\partial }{\partial y}\left(x{y}^{-1}\right)$
$=-x{y}^{-2}$
$=-\frac{x}{{y}^{2}}$
Result:
${z}_{x}=\frac{1}{{y}^{\prime }}$
${z}_{y}=-\frac{x}{{y}^{2}}$

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