Calculus Examples

$2 \cos (\frac{x}{2})$

Step 1

Find the first derivative.

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

$f' (x) = 2 \frac{d}{d x} (\cos (\frac{x}{2}))$

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

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f' (x) = 2 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (\frac{x}{2}))$

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$f' (x) = 2 (- \sin (u) \frac{d}{d x} \frac{x}{2})$

Replace all occurrences of $u$ with $\frac{x}{2}$ .

$f' (x) = 2 (- \sin (\frac{x}{2}) \frac{d}{d x} \frac{x}{2})$

Differentiate.

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Multiply $- 1$ by $2$ .

$f' (x) = - 2 (\sin (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2}))$

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

$f' (x) = - 2 \sin (\frac{x}{2}) (\frac{1}{2} \cdot \frac{d}{d x} (x))$

Simplify terms.

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

$f' (x) = \frac{- 2}{2} \cdot \sin (\frac{x}{2}) \frac{d}{d x} (x)$

Combine $\frac{- 2}{2}$ and $\sin (\frac{x}{2})$ .

$f' (x) = \frac{- 2 \sin (\frac{x}{2})}{2} \cdot \frac{d}{d x} (x)$

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

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Factor $2$ out of $- 2 \sin (\frac{x}{2})$ .

$f' (x) = \frac{2 (- \sin (\frac{x}{2}))}{2} \cdot \frac{d}{d x} (x)$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f' (x) = \frac{2 (- \sin (\frac{x}{2}))}{2 (1)} \cdot \frac{d}{d x} (x)$

Cancel the common factor.

$f' (x) = \frac{2 (- \sin (\frac{x}{2}))}{2 \cdot 1} \cdot \frac{d}{d x} (x)$

Rewrite the expression.

$f' (x) = \frac{- \sin (\frac{x}{2})}{1} \cdot \frac{d}{d x} (x)$

Divide $- \sin (\frac{x}{2})$ by $1$ .

$f' (x) = - \sin (\frac{x}{2}) \frac{d}{d x} x$

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

$f' (x) = - \sin (\frac{x}{2}) \cdot 1$

Multiply $- 1$ by $1$ .

$f' (x) = - \sin (\frac{x}{2})$

Step 2

Find the second derivative.

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

$f'' (x) = - \frac{d}{d x} \sin (\frac{x}{2})$

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

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f'' (x) = - (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (\frac{x}{2}))$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$f'' (x) = - (\cos (u) \frac{d}{d x} (\frac{x}{2}))$

Replace all occurrences of $u$ with $\frac{x}{2}$ .

$f'' (x) = - (\cos (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2}))$

Differentiate.

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

$f'' (x) = - \cos (\frac{x}{2}) (\frac{1}{2} \cdot \frac{d}{d x} (x))$

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

$f'' (x) = - \frac{\cos (\frac{x}{2})}{2} \cdot \frac{d}{d x} x$

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

$f'' (x) = - \frac{\cos (\frac{x}{2})}{2} \cdot 1$

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{\cos (\frac{x}{2})}{2}$

Step 3

Find the third derivative.

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

$f''' (x) = - \frac{1}{2} \cdot \frac{d}{d x} \cos (\frac{x}{2})$

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

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f''' (x) = - \frac{1}{2} \cdot (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (\frac{x}{2}))$

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$f''' (x) = - \frac{1}{2} \cdot (- \sin (u) \frac{d}{d x} \frac{x}{2})$

Replace all occurrences of $u$ with $\frac{x}{2}$ .

$f''' (x) = - \frac{1}{2} \cdot (- \sin (\frac{x}{2}) \frac{d}{d x} \frac{x}{2})$

Differentiate.

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Multiply $- 1$ by $- 1$ .

$f''' (x) = 1 (\frac{1}{2}) \cdot (\sin (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2}))$

Combine fractions.

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Multiply $\frac{1}{2}$ by $1$ .

$f''' (x) = \frac{1}{2} \cdot (\sin (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2}))$

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

$f''' (x) = \frac{\sin (\frac{x}{2})}{2} \cdot \frac{d}{d x} (\frac{x}{2})$

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

$f''' (x) = \frac{\sin (\frac{x}{2})}{2} \cdot (\frac{1}{2} \cdot \frac{d}{d x} (x))$

Combine fractions.

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Multiply $\frac{1}{2}$ by $\frac{\sin (\frac{x}{2})}{2}$ .

$f''' (x) = \frac{\sin (\frac{x}{2})}{2 \cdot 2} \cdot \frac{d}{d x} (x)$

Multiply $2$ by $2$ .

$f''' (x) = \frac{\sin (\frac{x}{2})}{4} \cdot \frac{d}{d x} (x)$

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

$f''' (x) = \frac{\sin (\frac{x}{2})}{4} \cdot 1$

Multiply $\frac{\sin (\frac{x}{2})}{4}$ by $1$ .

$f''' (x) = \frac{\sin (\frac{x}{2})}{4}$

Step 4

Find the fourth derivative.

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

$f^{4} (x) = \frac{1}{4} \cdot \frac{d}{d x} (\sin (\frac{x}{2}))$

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

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f^{4} (x) = \frac{1}{4} \cdot (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (\frac{x}{2}))$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$f^{4} (x) = \frac{1}{4} \cdot (\cos (u) \frac{d}{d x} (\frac{x}{2}))$

Replace all occurrences of $u$ with $\frac{x}{2}$ .

$f^{4} (x) = \frac{1}{4} \cdot (\cos (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2}))$

Differentiate.

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Combine $\cos (\frac{x}{2})$ and $\frac{1}{4}$ .

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{4} \cdot \frac{d}{d x} (\frac{x}{2})$

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

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{4} \cdot (\frac{1}{2} \cdot \frac{d}{d x} (x))$

Combine fractions.

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Multiply $\frac{1}{2}$ by $\frac{\cos (\frac{x}{2})}{4}$ .

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{2 \cdot 4} \cdot \frac{d}{d x} (x)$

Multiply $2$ by $4$ .

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{8} \cdot \frac{d}{d x} (x)$

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

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{8} \cdot 1$

Multiply $\frac{\cos (\frac{x}{2})}{8}$ by $1$ .

$f^{4} (x) = \frac{\cos (\frac{x}{2})}{8}$