Q. 60. - 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 $F$.

$g=\frac{1}{xy}-\sin x,h(x,y)=-\frac{\ln x}{y^2}$

Step-by-Step Solution

Answer

The required answer is $F(x,y)=\frac{\ln x}{y}+\cos x+C$

Step 1: Given information

Think about,

$g=\frac{1}{xy}-\sin x,h=-\frac{\ln x}{y^2}$

Then,

$g_y=-\frac{1}{x{y}^2}=h_x$

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

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

Think about,

$$\begin{gathered} \int (\frac{1}{xy}-\sin x)dx=\frac{\ln x}{y}+\cos x+q\left(y\right) \\ =F(x,y) \end{gathered}$$

Think about,

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

Suppose,

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

Hence, $F(x,y)=\frac{\ln x}{y}+\cos x+C$