Calculus Examples

$y = \frac{1}{4} \tan (2 x + 1)$

Step 1

Find the first derivative.

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Step 1.1

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{1}{4} \cdot \tan (2 x + 1)$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [\tan (2 x + 1)]$ .

$\frac{1}{4} \frac{d}{d x} [\tan (2 x + 1)]$

Step 1.2

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 $g (x) = 2 x + 1$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$\frac{1}{4} (\frac{d}{d u} [\tan (u)] \frac{d}{d x} [2 x + 1])$

Step 1.2.2

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

$\frac{1}{4} (\sec^{2} (u) \frac{d}{d x} [2 x + 1])$

Step 1.2.3

Replace all occurrences of $u$ with $2 x + 1$ .

$\frac{1}{4} (\sec^{2} (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 1.3

Differentiate.

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Step 1.3.1

Combine $\sec^{2} (2 x + 1)$ and $\frac{1}{4}$ .

$\frac{\sec^{2} (2 x + 1)}{4} \frac{d}{d x} [2 x + 1]$

Step 1.3.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{\sec^{2} (2 x + 1)}{4} (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])$

Step 1.3.3

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

$\frac{\sec^{2} (2 x + 1)}{4} (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 1.3.4

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

$\frac{\sec^{2} (2 x + 1)}{4} (2 \cdot 1 + \frac{d}{d x} [1])$

Step 1.3.5

Multiply $2$ by $1$ .

$\frac{\sec^{2} (2 x + 1)}{4} (2 + \frac{d}{d x} [1])$

Step 1.3.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{\sec^{2} (2 x + 1)}{4} (2 + 0)$

Step 1.3.7

Simplify terms.

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Step 1.3.7.1

Add $2$ and $0$ .

$\frac{\sec^{2} (2 x + 1)}{4} \cdot 2$

Step 1.3.7.2

Combine $\frac{\sec^{2} (2 x + 1)}{4}$ and $2$ .

$\frac{\sec^{2} (2 x + 1) \cdot 2}{4}$

Step 1.3.7.3

Move $2$ to the left of $\sec^{2} (2 x + 1)$ .

$\frac{2 \cdot \sec^{2} (2 x + 1)}{4}$

Step 1.3.7.4

Cancel the common factor of $2$ and $4$ .

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Step 1.3.7.4.1

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

$\frac{2 (\sec^{2} (2 x + 1))}{4}$

Step 1.3.7.4.2

Cancel the common factors.

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Step 1.3.7.4.2.1

Factor $2$ out of $4$ .

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

Step 1.3.7.4.2.2

Cancel the common factor.

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

Step 1.3.7.4.2.3

Rewrite the expression.

$f' (x) = \frac{\sec^{2} (2 x + 1)}{2}$

Step 2

Find the second derivative.

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Step 2.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{\sec^{2} (2 x + 1)}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\sec^{2} (2 x + 1)]$ .

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sec (2 x + 1)$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $\sec (2 x + 1)$ .

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

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

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

Step 2.2.3

Replace all occurrences of $u_{1}$ with $\sec (2 x + 1)$ .

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

Step 2.3

Simplify terms.

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Step 2.3.1

Combine $2$ and $\frac{1}{2}$ .

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

Step 2.3.2

Combine $\sec (2 x + 1)$ and $\frac{2}{2}$ .

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

Step 2.3.3

Move $2$ to the left of $\sec (2 x + 1)$ .

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

Step 2.3.4

Cancel the common factor of $2$ .

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Step 2.3.4.1

Cancel the common factor.

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

Step 2.3.4.2

Divide $\sec (2 x + 1)$ by $1$ .

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

Step 2.4

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 $g (x) = 2 x + 1$ .

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

Add $1$ and $1$ .

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

Step 2.9

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 2.10

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

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

Step 2.11

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

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

Step 2.12

Multiply $2$ by $1$ .

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

Step 2.13

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\sec^{2} (2 x + 1) \tan (2 x + 1) (2 + 0)$

Step 2.14

Simplify the expression.

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Step 2.14.1

Add $2$ and $0$ .

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

Step 2.14.2

Move $2$ to the left of $\sec^{2} (2 x + 1) \tan (2 x + 1)$ .

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