91Ó°ÊÓ

$∇f(x,y)=(2x+\cos x\cos y)\mathrm{i}-\sin x\sin y\mathrm{j}$

Consider the gradient

$∇f(x,y)=(2x+\cos x\cos y)\mathrm{i}-\sin x\sin y\mathrm{j}……\text{. (1)}$

The goal is to deduce the function from the gradient.

Rewrite ( 1) as

$$\begin{gathered} ∇f(x,y)=(2x+\cos x\cos y)\mathrm{i}-\sin x\sin y\mathrm{j} \\ {f}_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}=(2x+\cos x\cos y)\mathrm{i}-\sin x\sin y\mathrm{j} \end{gathered}$$

Equate both sides

$f_x(x,y)=2x+\cos x\cos y……\text{(2)}$

And

$f_y(x,y)=-\sin x\sin y…….\left(3\right)$

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

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

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

$$\begin{gathered} f_{xy}(x,y)=2\frac{∂}{∂y}x+\frac{∂}{∂y}\left(\cos x\cos y\right) \\ =2·0+\cos x\frac{∂}{∂y}\cos y \\ =\cos x(-\sin y) \\ =-\cos x\sin y \end{gathered}$$

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

$$\begin{gathered} f_{yx}(x,y)=\frac{∂}{∂x}(-\sin x\sin y) \\ =-\sin y\frac{∂}{∂x}\sin x \\ =-\sin y\cos x \end{gathered}$$

Since $f_{xy}(x,y)=-\sin y\cos x=f_{yx}(x,y)$so the function exists.

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

$$\begin{gathered} f(x,y)=∫(-\sin x\sin y)dy+q\left(x\right) \\ =-\sin x∫\sin ydy+q\left(x\right) \\ =-\sin x(-\cos y)+q\left(x\right) \\ =\sin x\cos y+q\left(x\right)…… \end{gathered}$$

where $q\left(x\right)$is an arbitrary function.

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

$\frac{∂}{∂x}f(x,y)=\frac{∂}{∂x}\sin x\cos y+\frac{∂}{∂x}q\left(x\right)$$$\begin{gathered} f_x(x,y)=\cos y\frac{∂}{∂x}\sin x+q^{\text{'}}\left(x\right) \\ =\cos y\cos x+q^{\text{'}}\left(x\right) \\ 2x+\cos x\cos y=\cos y\cos x+q^{\text{'}}\left(x\right)\left[\text{From}\left(2\right);f_x(x,y)=2x+\cos x\cos y\text{]}\right. \end{gathered}$$$2x=q^{\text{'}}\left(x\right)$$$\begin{gathered} q^{\text{'}}\left(x\right)=2x⋯⋯\left(5\right) \\ \text{Integrate}\left(5\right)\text{with respect to}y. \\ ∫q^{\text{'}}\left(x\right)dx=∫2xdx+C \\ q\left(x\right)=2∫xdx+C \\ =2·\frac{x^2}{2}+C \\ =x^2+C \end{gathered}$$

where $C$is the constant of integration.

Put $q\left(x\right)=x^2+C$ in $(4)$

$f(x,y)=\sin x\cos y+x^2+C\text{.}$

As a result, the necessary function is $f(x,y)=\sin x\cos y+x^2+C\text{.}$