91Ó°ÊÓ

$∇f(x,y)=\left(\frac{1}{y}-\frac{y}{x^2}\right)\mathrm{i}+\left(\frac{1}{x}-\frac{x}{y^2}\right)\mathrm{j}$

Consider the gradient

$∇f(x,y)=\left(\frac{1}{y}-\frac{y}{x^2}\right)\mathrm{i}+\left(\frac{1}{x}-\frac{x}{y^2}\right)\mathrm{j}$

The goal is to deduce the function from the gradient.

Rewrite $\left(1\right)$as

$$\begin{gathered} ∇f(x,y)=\left(\frac{1}{y}-\frac{y}{x^2}\right)\mathrm{i}+\left(\frac{1}{x}-\frac{x}{y^2}\right)\mathrm{j} \\ {f}_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}=\left(\frac{1}{y}-\frac{y}{x^2}\right)\mathrm{i}+\left(\frac{1}{x}-\frac{x}{y^2}\right)\mathrm{j} \end{gathered}$$

Equate both sides

$f_x(x,y)=\left(\frac{1}{y}-\frac{y}{x^2}\right)……\left(2\right)$

And

$f_y(x,y)=\left(\frac{1}{x}-\frac{x}{y^2}\right)……\left(3\right)$

Check to see if the function is available. If there is a function, it exists.

$f_{xy}(x,y)=f_{yx}(x,y)$

Now, find $f_{xy}(x,y)$ by differentiating $f_x(x,y)$ partially with respect to $y$

Also, find $f_{yx}(x,y)$ by differentiating $f_y(x,y)$ partially with respect to $x$

Since $f_{xy}(x,y)=-\frac{1}{x^2}-\frac{1}{y^2}=f_{yx}(x,y)$so the function exists.

Integrate $\left(3\right)$with respect to $y$

$f(x,y)=∫\left(\frac{1}{x}-\frac{x}{y^2}\right)dy+q\left(x\right)$where $q\left(x\right)$is an arbitrary function

$$\begin{gathered} =∫\frac{1}{x}dy-∫\frac{x}{y^2}dy+q\left(x\right) \\ =\frac{1}{x}∫dy-x∫\frac{1}{y^2}dy+q\left(x\right) \\ =\frac{y}{x}-x·\frac{y^{-2+1}}{(-2+1)}+q\left(x\right) \\ =\frac{y}{x}-x·\frac{y^{-1}}{(-1)}+q\left(x\right) \\ =\frac{y}{x}+\frac{x}{y}+q\left(x\right)…..\left(4\right) \end{gathered}$$

Next, find $q\left(x\right)$to partially differentiate $\left(4\right)$with regard to $x$

$$\begin{gathered} \frac{∂}{∂x}f(x,y)=\frac{∂}{∂x}\left(\frac{y}{x}+\frac{x}{y}\right)+\frac{∂}{∂x}q\left(x\right) \\ {f}_x(x,y)=\left(\frac{∂}{∂x}\frac{y}{x}+\frac{∂}{∂x}\frac{x}{y}\right)+q^{\text{'}}\left(x\right) \\ =\left(y\frac{∂}{∂x}\frac{1}{x}+\frac{1}{y}\frac{∂}{∂x}x\right)+q^{\text{'}}\left(x\right) \end{gathered}$$

$$\begin{gathered} =\left(y(-1)x^{-1-1}+\frac{1}{y}\right)+q^{\text{'}}\left(x\right) \\ =\left(-y{x}^{-2}+\frac{1}{y}\right)+q^{\text{'}}\left(x\right) \\ =-\frac{y}{x^2}+\frac{1}{y}+q^{\text{'}}\left(x\right) \\ \text{Comment} \\ ∫q^{\text{'}}\left(x\right)dx=∫0dx+C \\ \text{Integrate}\left(5\right)\text{with respect to}x. \\ \frac{1}{y}-\frac{y}{x^2}=-\frac{y}{x^2}+\frac{1}{y}+q^{\text{'}}\left(x\right)\left[\text{From}\left(2\right);f_x(x,y)=\left(\frac{1}{y}-\frac{y}{x^2}\right)\right] \\ \frac{1}{y}-\frac{y}{x^2}+\frac{y}{x^2}-\frac{1}{y}=q^{\text{'}}\left(x\right) \\ {q}^{\text{'}}\left(x\right)=0 \end{gathered}$$

Integrate $\left(5\right)$with respect to $x$

$$\begin{gathered} ∫q^{\text{'}}\left(x\right)dx=∫0dx+C \\ q\left(x\right)=0+C \\ =C \end{gathered}$$

where $C$ is the constant of integration.

Put $q\left(x\right)=C$ in $(4)$

$f(x,y)=\frac{y}{x}+\frac{x}{y}+C$

Therefore, the required function is

$f(x,y)=\frac{y}{x}+\frac{x}{y}+C$