Calculus Examples

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

Step 1

Multiply $x$ by $\tan (y)$ .

$y = x \tan (y)$

Step 2

Differentiate both sides of the equation.

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

Step 3

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 4

Differentiate the right side of the equation.

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Step 4.1

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 $g (x) = \tan (y)$ .

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

Step 4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = y$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $y$ .

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

Step 4.2.2

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

$x (\sec^{2} (u) \frac{d}{d x} [y]) + \tan (y) \frac{d}{d x} [x]$

Step 4.2.3

Replace all occurrences of $u$ with $y$ .

$x (\sec^{2} (y) \frac{d}{d x} [y]) + \tan (y) \frac{d}{d x} [x]$

Step 4.3

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

$x \sec^{2} (y) y' + \tan (y) \frac{d}{d x} [x]$

Step 4.4

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

$x \sec^{2} (y) y' + \tan (y) \cdot 1$

Step 4.5

Multiply $\tan (y)$ by $1$ .

$x \sec^{2} (y) y' + \tan (y)$

Step 5

Reform the equation by setting the left side equal to the right side.

$y' = x \sec^{2} (y) y' + \tan (y)$

Step 6

Solve for $y'$ .

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Step 6.1

Simplify the right side.

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Step 6.1.1

Reorder factors in $x \sec^{2} (y) y' + \tan (y)$ .

$y' = x y' \sec^{2} (y) + \tan (y)$

Step 6.2

Subtract $x y' \sec^{2} (y)$ from both sides of the equation.

$y' - x y' \sec^{2} (y) = \tan (y)$

Step 6.3

Factor $y'$ out of $y' - x y' \sec^{2} (y)$ .

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Step 6.3.1

Factor $y'$ out of ${y'}^{1}$ .

$y' \cdot 1 - x y' \sec^{2} (y) = \tan (y)$

Step 6.3.2

Factor $y'$ out of $- x y' \sec^{2} (y)$ .

$y' \cdot 1 + y' (- x \sec^{2} (y)) = \tan (y)$

Step 6.3.3

Factor $y'$ out of $y' \cdot 1 + y' (- x \sec^{2} (y))$ .

$y' (1 - x \sec^{2} (y)) = \tan (y)$

Step 6.4

Divide each term in $y' (1 - x \sec^{2} (y)) = \tan (y)$ by $1 - x \sec^{2} (y)$ and simplify.

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Step 6.4.1

Divide each term in $y' (1 - x \sec^{2} (y)) = \tan (y)$ by $1 - x \sec^{2} (y)$ .

$\frac{y' (1 - x \sec^{2} (y))}{1 - x \sec^{2} (y)} = \frac{\tan (y)}{1 - x \sec^{2} (y)}$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Cancel the common factor of $1 - x \sec^{2} (y)$ .

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Step 6.4.2.1.1

Cancel the common factor.

$\frac{y' (1 - x \sec^{2} (y))}{1 - x \sec^{2} (y)} = \frac{\tan (y)}{1 - x \sec^{2} (y)}$

Step 6.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{\tan (y)}{1 - x \sec^{2} (y)}$

Step 7

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{\tan (y)}{1 - x \sec^{2} (y)}$