Calculus Examples

$- \frac{1}{4} \sin (2 x)$

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 $- \frac{1}{4} \sin (2 x)$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [\sin (2 x)]$ .

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

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

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

$- \frac{\cos (2 x)}{4} \frac{d}{d x} [2 x]$

Step 1.3.2

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

$- \frac{\cos (2 x)}{4} (2 \frac{d}{d x} [x])$

Step 1.3.3

Simplify terms.

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Step 1.3.3.1

Multiply $2$ by $- 1$ .

$- 2 \frac{\cos (2 x)}{4} \frac{d}{d x} [x]$

Step 1.3.3.2

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

$\frac{- 2 \cos (2 x)}{4} \frac{d}{d x} [x]$

Step 1.3.3.3

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

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Step 1.3.3.3.1

Factor $2$ out of $- 2 \cos (2 x)$ .

$\frac{2 (- \cos (2 x))}{4} \frac{d}{d x} [x]$

Step 1.3.3.3.2

Cancel the common factors.

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Step 1.3.3.3.2.1

Factor $2$ out of $4$ .

$\frac{2 (- \cos (2 x))}{2 (2)} \frac{d}{d x} [x]$

Step 1.3.3.3.2.2

Cancel the common factor.

$\frac{2 (- \cos (2 x))}{2 \cdot 2} \frac{d}{d x} [x]$

Step 1.3.3.3.2.3

Rewrite the expression.

$\frac{- \cos (2 x)}{2} \frac{d}{d x} [x]$

Step 1.3.3.4

Move the negative in front of the fraction.

$- \frac{\cos (2 x)}{2} \frac{d}{d x} [x]$

Step 1.3.4

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

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

Step 1.3.5

Multiply $- 1$ by $1$ .

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

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 $- \frac{\cos (2 x)}{2}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [\cos (2 x)]$ .

$- \frac{1}{2} \frac{d}{d x} [\cos (2 x)]$

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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

$- \frac{1}{2} (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.2

Combine fractions.

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

$\frac{\sin (2 x)}{2} \frac{d}{d x} [2 x]$

Step 2.3.3

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

$\frac{\sin (2 x)}{2} (2 \frac{d}{d x} [x])$

Step 2.3.4

Simplify terms.

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Step 2.3.4.1

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

$\frac{2 \sin (2 x)}{2} \frac{d}{d x} [x]$

Step 2.3.4.2

Cancel the common factor of $2$ .

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Step 2.3.4.2.1

Cancel the common factor.

$\frac{2 \sin (2 x)}{2} \frac{d}{d x} [x]$

Step 2.3.4.2.2

Divide $\sin (2 x)$ by $1$ .

$\sin (2 x) \frac{d}{d x} [x]$

Step 2.3.5

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

$\sin (2 x) \cdot 1$

Step 2.3.6

Multiply $\sin (2 x)$ by $1$ .

$f'' (x) = \sin (2 x)$