Calculus Examples

$\frac{d y}{d x} = - \frac{x}{y}$

Write $\frac{y}{x} + \frac{x}{y} = 0$ as a function.

$f (y) = \frac{y}{x} + \frac{x}{y}$

Add $\frac{x}{y}$ to both sides of the equation.

$\frac{d y}{d x} + \frac{x}{y} = 0$

Cancel the common factor of $d$ .

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Cancel the common factor.

$\frac{d y}{d x} + \frac{x}{y} = 0$

Rewrite the expression.

$\frac{y}{x} + \frac{x}{y} = 0$

The function $F (y)$ can be found by finding the indefinite integral of the derivative $f (y)$ .

$F (y) = \int f (y) d y$

Set up the integral to solve.

$F (y) = \int \frac{y}{x} + \frac{x}{y} d y$

Split the single integral into multiple integrals.

$\int \frac{y}{x} d y + \int \frac{x}{y} d y$

Since $\frac{1}{x}$ is constant with respect to $y$ , move $\frac{1}{x}$ out of the integral.

$\frac{1}{x} \int y d y + \int \frac{x}{y} d y$

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

$\frac{1}{x} (\frac{1}{2} y^{2} + C) + \int \frac{x}{y} d y$

Since $x$ is constant with respect to $y$ , move $x$ out of the integral.

$\frac{1}{x} (\frac{1}{2} y^{2} + C) + x \int \frac{1}{y} d y$

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$\frac{1}{x} (\frac{1}{2} y^{2} + C) + x (\ln (| y |) + C)$

Simplify.

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Simplify.

$\frac{1}{2} \cdot \frac{1}{x} y^{2} + x \ln (| y |) + C$

Simplify.

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Multiply $\frac{1}{2}$ by $\frac{1}{x}$ .

$\frac{1}{2 x} y^{2} + x \ln (| y |) + C$

Combine $\frac{1}{2 x}$ and $y^{2}$ .

$\frac{y^{2}}{2 x} + x \ln (| y |) + C$

The answer is the antiderivative of the function $f (y) = \frac{y}{x} + \frac{x}{y}$ .

$F (y) =$ $\frac{y^{2}}{2 x} + x \ln (| y |) + C$