Calculus Examples

$y = \sec (θ) \tan (θ)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} (\sec (θ) \tan (θ))$

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

To apply the Chain Rule, set $u_{1}$ as $θ$ .

$\sec (θ) (\frac{d}{d u_{1}} [\tan (u_{1})] \frac{d}{d y} [θ]) + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.2.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

$\sec (θ) (\sec^{2} (u_{1}) \frac{d}{d y} [θ]) + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.2.3

Replace all occurrences of $u_{1}$ with $θ$ .

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

Step 3.3

Raise $\sec (θ)$ to the power of $1$ .

$\sec^{2} (θ) \sec^{1} (θ) \frac{d}{d y} [θ] + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${\sec (θ)}^{2 + 1} \frac{d}{d y} [θ] + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.5

Add $2$ and $1$ .

$\sec^{3} (θ) \frac{d}{d y} [θ] + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.6

Rewrite $\frac{d}{d y} [θ]$ as $θ'$ .

$\sec^{3} (θ) θ' + \tan (θ) \frac{d}{d y} [\sec (θ)]$

Step 3.7

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

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Step 3.7.1

To apply the Chain Rule, set $u_{2}$ as $θ$ .

$\sec^{3} (θ) θ' + \tan (θ) (\frac{d}{d u_{2}} [\sec (u_{2})] \frac{d}{d y} [θ])$

Step 3.7.2

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

$\sec^{3} (θ) θ' + \tan (θ) (\sec (u_{2}) \tan (u_{2}) \frac{d}{d y} [θ])$

Step 3.7.3

Replace all occurrences of $u_{2}$ with $θ$ .

$\sec^{3} (θ) θ' + \tan (θ) (\sec (θ) \tan (θ) \frac{d}{d y} [θ])$

Step 3.8

Raise $\tan (θ)$ to the power of $1$ .

$\sec^{3} (θ) θ' + \tan^{1} (θ) \tan (θ) (\sec (θ) \frac{d}{d y} [θ])$

Step 3.9

Raise $\tan (θ)$ to the power of $1$ .

$\sec^{3} (θ) θ' + \tan^{1} (θ) \tan^{1} (θ) (\sec (θ) \frac{d}{d y} [θ])$

Step 3.10

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sec^{3} (θ) θ' + {\tan (θ)}^{1 + 1} (\sec (θ) \frac{d}{d y} [θ])$

Step 3.11

Add $1$ and $1$ .

$\sec^{3} (θ) θ' + \tan^{2} (θ) (\sec (θ) \frac{d}{d y} [θ])$

Step 3.12

Rewrite $\frac{d}{d y} [θ]$ as $θ'$ .

$\sec^{3} (θ) θ' + \tan^{2} (θ) \sec (θ) θ'$

Step 4

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

$1 = \sec^{3} (θ) θ' + \tan^{2} (θ) \sec (θ) θ'$

Step 5

Solve for $θ'$ .

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Step 5.1

Rewrite the equation as $\sec^{3} (θ) θ' + \tan^{2} (θ) \sec (θ) θ' = 1$ .

$\sec^{3} (θ) θ' + \tan^{2} (θ) \sec (θ) θ' = 1$

Step 5.2

Simplify the left side.

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Step 5.2.1

Reorder factors in $\sec^{3} (θ) θ' + \tan^{2} (θ) \sec (θ) θ'$ .

$θ' \sec^{3} (θ) + θ' \tan^{2} (θ) \sec (θ) = 1$

Step 5.3

Factor $θ' \sec (θ)$ out of $θ' \sec^{3} (θ) + θ' \tan^{2} (θ) \sec (θ)$ .

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Step 5.3.1

Factor $θ' \sec (θ)$ out of $θ' \sec^{3} (θ)$ .

$θ' \sec (θ) \sec^{2} (θ) + θ' \tan^{2} (θ) \sec (θ) = 1$

Step 5.3.2

Factor $θ' \sec (θ)$ out of $θ' \tan^{2} (θ) \sec (θ)$ .

$θ' \sec (θ) \sec^{2} (θ) + θ' \sec (θ) \tan^{2} (θ) = 1$

Step 5.3.3

Factor $θ' \sec (θ)$ out of $θ' \sec (θ) \sec^{2} (θ) + θ' \sec (θ) \tan^{2} (θ)$ .

$θ' \sec (θ) (\sec^{2} (θ) + \tan^{2} (θ)) = 1$

Step 5.4

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

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Step 5.4.1

Divide each term in $θ' \sec (θ) (\sec^{2} (θ) + \tan^{2} (θ)) = 1$ by $\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))$ .

$\frac{θ' \sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))} = \frac{1}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $\sec (θ)$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{θ' \sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))} = \frac{1}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{θ' (\sec^{2} (θ) + \tan^{2} (θ))}{\sec^{2} (θ) + \tan^{2} (θ)} = \frac{1}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}$

Step 5.4.2.2

Cancel the common factor of $\sec^{2} (θ) + \tan^{2} (θ)$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{θ' (\sec^{2} (θ) + \tan^{2} (θ))}{\sec^{2} (θ) + \tan^{2} (θ)} = \frac{1}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}$

Step 5.4.2.2.2

Divide $θ'$ by $1$ .

$θ' = \frac{1}{\sec (θ) (\sec^{2} (θ) + \tan^{2} (θ))}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Separate fractions.

$θ' = \frac{1}{\sec^{2} (θ) + \tan^{2} (θ)} \cdot \frac{1}{\sec (θ)}$

Step 5.4.3.2

Rewrite $\sec (θ)$ in terms of sines and cosines.

$θ' = \frac{1}{\sec^{2} (θ) + \tan^{2} (θ)} \cdot \frac{1}{\frac{1}{\cos (θ)}}$

Step 5.4.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (θ)}$ .

$θ' = \frac{1}{\sec^{2} (θ) + \tan^{2} (θ)} (1 \cos (θ))$

Step 5.4.3.4

Multiply $\cos (θ)$ by $1$ .

$θ' = \frac{1}{\sec^{2} (θ) + \tan^{2} (θ)} \cos (θ)$

Step 5.4.3.5

Combine $\frac{1}{\sec^{2} (θ) + \tan^{2} (θ)}$ and $\cos (θ)$ .

$θ' = \frac{\cos (θ)}{\sec^{2} (θ) + \tan^{2} (θ)}$

Step 6

Replace $θ'$ with $\frac{d θ}{d y}$ .

$\frac{d θ}{d y} = \frac{\cos (θ)}{\sec^{2} (θ) + \tan^{2} (θ)}$