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How to check whether P dx + Q dy is an exact differential

The test: compare ∂P/∂y with ∂Q/∂x

An expression $P (x, y) d x + Q (x, y) d y$ is the exact differential of some function $u (x, y)$ exactly when the two mixed partial derivatives agree:

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

So the whole job is: read off $P$ and $Q$ , differentiate $P$ with respect to $y$ (treating $x$ as a constant), differentiate $Q$ with respect to $x$ (treating $y$ as a constant), and compare. If the two results are identical on a region without holes, the expression is exact; if they differ anywhere, it is not.

Problem

Check whether the expression

\frac{2 x ( 1 - e ^{y} )}{( 1 + x ^{2} ) ^{2}} d x + (\frac{e ^{y}}{1 + x ^{2}} + 1) d y

is the exact differential of a function of two variables.

Answer

Yes - the expression is an exact differential, because

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = - \frac{2 x e ^{y}}{( 1 + x ^{2} ) ^{2}}

Checked by hand: both derivatives were recomputed and they agree.

Step-by-step solution

1. Read off P and Q

The expression has the form $P (x, y) d x + Q (x, y) d y$ , so the coefficients of the differentials are

P (x, y) = \frac{2 x ( 1 - e ^{y} )}{( 1 + x ^{2} ) ^{2}}

Q (x, y) = \frac{e ^{y}}{1 + x ^{2}} + 1

2. Differentiate P with respect to y

Here $x$ is treated as a constant, so the whole factor $\frac{2 x}{( 1 + x ^{2} ) ^{2}}$ is a constant and only $1 - e^{y}$ is differentiated:

\frac{\partial P}{\partial y} = \frac{2 x}{( 1 + x ^{2} ) ^{2}} \cdot (- e^{y}) = - \frac{2 x e ^{y}}{( 1 + x ^{2} ) ^{2}}

3. Differentiate Q with respect to x

Now $y$ is the constant, the term $+ 1$ disappears, and the chain rule applies to $(1 + x^{2})^{- 1}$ :

\frac{\partial Q}{\partial x} = e^{y} \cdot \frac{\partial}{\partial x} (\frac{1}{1 + x ^{2}}) = e^{y} \cdot (- \frac{2 x}{( 1 + x ^{2} ) ^{2}}) = - \frac{2 x e ^{y}}{( 1 + x ^{2} ) ^{2}}

4. Compare

\frac{\partial P}{\partial y} = - \frac{2 x e ^{y}}{( 1 + x ^{2} ) ^{2}}, \frac{\partial Q}{\partial x} = - \frac{2 x e ^{y}}{( 1 + x ^{2} ) ^{2}}

The two derivatives coincide for every $x$ and $y$ (the denominator $1 + x^{2}$ never vanishes, so the region is the whole plane and has no holes). The condition is satisfied, so the expression is an exact differential.

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