Calculus Examples

$f (x) = \tan^{-1} (2 x)$

Step 1

Find the first derivative.

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Step 1.1

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

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Step 1.1.1

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

$\frac{d}{d u} [\tan^{-1} (u)] \frac{d}{d x} [2 x]$

Step 1.1.2

The derivative of $\tan^{-1} (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

$\frac{1}{1 + u^{2}} \frac{d}{d x} [2 x]$

Step 1.1.3

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

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

Step 1.2

Differentiate.

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Step 1.2.1

Factor $2$ out of $2 x$ .

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

Step 1.2.2

Simplify the expression.

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Step 1.2.2.1

Apply the product rule to $2 (x)$ .

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

Step 1.2.2.2

Raise $2$ to the power of $2$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Multiply $\frac{2}{1 + 4 x^{2}}$ by $1$ .

$\frac{2}{1 + 4 x^{2}}$

Step 1.2.6.2

Reorder terms.

$f' (x) = \frac{2}{4 x^{2} + 1}$

Step 2

Find the second derivative.

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Step 2.1

Differentiate using the Constant Multiple Rule.

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Step 2.1.1

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

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

Step 2.1.2

Rewrite $\frac{1}{4 x^{2} + 1}$ as ${(4 x^{2} + 1)}^{- 1}$ .

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

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 $f (x) = x^{- 1}$ and $g (x) = 4 x^{2} + 1$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $4 x^{2} + 1$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $4 x^{2} + 1$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $2$ .

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

Step 2.3.2

By the Sum Rule, the derivative of $4 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$ .

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

Step 2.3.3

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $2$ by $4$ .

$- 2 {(4 x^{2} + 1)}^{- 2} (8 x + \frac{d}{d x} [1])$

Step 2.3.6

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

$- 2 {(4 x^{2} + 1)}^{- 2} (8 x + 0)$

Step 2.3.7

Simplify the expression.

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Step 2.3.7.1

Add $8 x$ and $0$ .

$- 2 {(4 x^{2} + 1)}^{- 2} (8 x)$

Step 2.3.7.2

Multiply $8$ by $- 2$ .

$- 16 {(4 x^{2} + 1)}^{- 2} x$

Step 2.4

Simplify.

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Step 2.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 16 \frac{1}{{(4 x^{2} + 1)}^{2}} x$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 16$ and $\frac{1}{{(4 x^{2} + 1)}^{2}}$ .

$\frac{- 16}{{(4 x^{2} + 1)}^{2}} x$

Step 2.4.2.2

Move the negative in front of the fraction.

$- \frac{16}{{(4 x^{2} + 1)}^{2}} x$

Step 2.4.2.3

Combine $x$ and $\frac{16}{{(4 x^{2} + 1)}^{2}}$ .

$- \frac{x \cdot 16}{{(4 x^{2} + 1)}^{2}}$

Step 2.4.2.4

Move $16$ to the left of $x$ .

$f'' (x) = - \frac{16 x}{{(4 x^{2} + 1)}^{2}}$

Step 3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{16 x}{{(4 x^{2} + 1)}^{2}}$ .

$- \frac{16 x}{{(4 x^{2} + 1)}^{2}}$