William Cleghorn

Answered

2022-01-17

Compute the first-order partial derivatives.

$z=\frac{x}{y}$

Answer & Explanation

Toni Scott

Expert

2022-01-18Added 32 answers

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.

${z}_{y}=\frac{\partial}{\partial y}\left(\frac{x}{y}\right)$

$=x\frac{\partial}{\partial y}\left(\frac{1}{y}\right)$

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

$=x\cdot (-{y}^{-2})$

$=-\frac{x}{{y}^{2}}$

servidopolisxv

Expert

2022-01-16Added 27 answers

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:

$z}_{x}=\frac{1}{{y}^{\prime}$

$z}_{y}=-\frac{x}{{y}^{2}$

The first-order partial derivatives:

Result:

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