Calculus Examples

$\sec (2 x) \tan (2 x)$

Step 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 $f (x) = \sec (2 x)$ and $g (x) = \tan (2 x)$ .

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

Step 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 $g (x) = 2 x$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 3

Raise $\sec (2 x)$ to the power of $1$ .

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

Step 4

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

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

Step 5

Differentiate.

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Step 5.1

Add $2$ and $1$ .

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

Step 5.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 5.3

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

$\sec^{3} (2 x) (2 \cdot 1) + \tan (2 x) \frac{d}{d x} [\sec (2 x)]$

Step 5.4

Simplify the expression.

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Step 5.4.1

Multiply $2$ by $1$ .

$\sec^{3} (2 x) \cdot 2 + \tan (2 x) \frac{d}{d x} [\sec (2 x)]$

Step 5.4.2

Move $2$ to the left of $\sec^{3} (2 x)$ .

$2 \cdot \sec^{3} (2 x) + \tan (2 x) \frac{d}{d x} [\sec (2 x)]$

Step 6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x$ .

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 7

Raise $\tan (2 x)$ to the power of $1$ .

$2 \sec^{3} (2 x) + \tan^{1} (2 x) \tan (2 x) (\sec (2 x) \frac{d}{d x} [2 x])$

Step 8

Raise $\tan (2 x)$ to the power of $1$ .

$2 \sec^{3} (2 x) + \tan^{1} (2 x) \tan^{1} (2 x) (\sec (2 x) \frac{d}{d x} [2 x])$

Step 9

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

$2 \sec^{3} (2 x) + {\tan (2 x)}^{1 + 1} (\sec (2 x) \frac{d}{d x} [2 x])$

Step 10

Add $1$ and $1$ .

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

Step 11

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 12

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

$2 \sec^{3} (2 x) + \tan^{2} (2 x) \sec (2 x) (2 \cdot 1)$

Step 13

Simplify the expression.

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Step 13.1

Multiply $2$ by $1$ .

$2 \sec^{3} (2 x) + \tan^{2} (2 x) \sec (2 x) \cdot 2$

Step 13.2

Move $2$ to the left of $\tan^{2} (2 x) \sec (2 x)$ .

$2 \sec^{3} (2 x) + 2 \cdot (\tan^{2} (2 x) \sec (2 x))$

Step 13.3

Reorder terms.

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