Q. 61. - Calculus

Question

For each pair of functions in Exercises 59–62, use Theorem

12.24 to show that there is a function of two variables,

$F(x,y)$ such that $\frac{dF}{dx}=g(x,y)$ and $\frac{dF}{dy}=h(x,y)$ Then find $F$.

$g=\frac{y}{1+x^2{y}^2},h(x,y)=\frac{x}{1+x^2{y}^2}$

Step-by-Step Solution

Answer

The required answer is $F(x,y)={\tan }^{-1}\left(xy\right)+C$

Step 1: Given information

Think about,

$g=\frac{y}{1+x^2{y}^2},h(x,y)=\frac{x}{1+x^2{y}^2}$

Then,

$$\begin{gathered} g_y=\frac{\left(1+x^2{y}^2\right)\left(1\right)-y\left(2x^{2}y\right)}{{\left(1+x^2{y}^2\right)}^2} \\ =\frac{1-x^2{y}^2}{{\left(1+x^2{y}^2\right)}^2} \\ =h_x \end{gathered}$$

There is a function $F(x, y)$ based on the Theorem $12.24$

Step 2: The objective is to find F integration with respect to x

Think about,

$$\begin{gathered} \int \left(\frac{y}{1+x^2{y}^2}\right)dx=y\frac{1}{1}\frac{{\tan }^{-1}\left(\frac{yx}{1}\right)}{y}+q\left(y\right) \\ ={\tan }^{-1}\left(yx\right)+q\left(y\right) \\ =F(x, y) \end{gathered}$$

Think about,

$$\begin{gathered} \frac{d}{dy}\left(F\right(x,y\left)\right)=\frac{d}{dy}\left({\tan }^{-1}\right(yx)+q(y\left)\right) \\ =\frac{1}{1+{\left(yx\right)}^2}\left(x\right)+q^{\text{'}}\left(y\right) \end{gathered}$$

Suppose,

$$\begin{gathered} \frac{d}{dy}\left(F\right(x,y\left)\right)=h \\ \Rightarrow \frac{x}{1+x^2{y}^2}+q^{\text{'}}\left(y\right)=\frac{x}{1+x^2{y}^2} \\ \Rightarrow q^{\text{'}}\left(y\right)=0 \\ q\left(y\right)=C \end{gathered}$$

Hence, $F(x,y)={\tan }^{-1}\left(xy\right)+C$