Calculus Examples

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

Step 1

Check if the left side of the equation is the result of the derivative of the term $x y$ .

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y$ .

$x \frac{d}{d x} [y] + y \frac{d}{d x} [x]$

Step 1.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x y' + y \frac{d}{d x} [x]$

Step 1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x y' + y \cdot 1$

Step 1.4

Substitute $\frac{d y}{d x}$ for $y'$ .

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

Step 1.5

Reorder $y$ and $1$ .

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

Step 1.6

Multiply $y$ by $1$ .

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

Step 2

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [x y] = \tan (x)$

Step 3

Set up an integral on each side.

$\int \frac{d}{d x} [x y] d x = \int \tan (x) d x$

Step 4

Integrate the left side.

$x y = \int \tan (x) d x$

Step 5

The integral of $\tan (x)$ with respect to $x$ is $\ln (| \sec (x) |)$ .

$x y = \ln (| \sec (x) |) + C$

Step 6

Divide each term in $x y = \ln (| \sec (x) |) + C$ by $x$ and simplify.

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Step 6.1

Divide each term in $x y = \ln (| \sec (x) |) + C$ by $x$ .

$\frac{x y}{x} = \frac{\ln (| \sec (x) |)}{x} + \frac{C}{x}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $x$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{x y}{x} = \frac{\ln (| \sec (x) |)}{x} + \frac{C}{x}$

Step 6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (| \sec (x) |)}{x} + \frac{C}{x}$