For each pair of functions in Exercises 59–62, use Theorem12.24 to show that there is a function of two variables,f(x,y) such that role="math" localid="1653976646361" dFdx=g(x,y) and dFdy=h(x,y) Then find F.g=2xy,h(x,y)=-x2y2

The required answer is F(x,y)=x2y+C

Q. 62. - Calculus | StudyQuestionHub

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 $\frac{d F}{d x} = g (x, y)$ and $\frac{d F}{d y} = h (x, y)$ Then find $F$ .

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

Verified

Answer

The required answer is $F (x, y) = \frac{x^{2}}{y} + C$

1Step 1: Given information

Think about, $g = \frac{2 x}{y}, h = - \frac{x^{2}}{y^{2}}$

Then,

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

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

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

Think about,

$\int (\frac{2 x}{y}) d x = \frac{2}{y} \frac{x^{2}}{2} + q (y) = \frac{x^{2}}{y} + q (y) = F (x, y)$

Think about,

$\frac{d}{d y} (F (x, y)) = \frac{d}{d y} (\frac{x^{2}}{y} + q (y)) = - \frac{x^{2}}{y^{2}} + q^{'} (y)$

Suppose,

$\frac{d}{d y} (F (x, y)) = h \Rightarrow - \frac{x^{2}}{y^{2}} + q^{'} (y) = - \frac{x^{2}}{y^{2}} \Rightarrow q^{'} (y) = 0 q (y) = C$

Hence, $F (x, y) = \frac{x^{2}}{y} + C$