Calculus Examples

$f (x) = [\sin (3 x)]$

Step 1

Find the first derivative.

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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 $g (x) = 3 x$ .

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To apply the Chain Rule, set $u$ as $3 x$ .

$f' (x) = \frac{d}{d u} (\sin (u)) \frac{d}{d x} (3 x)$

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

$f' (x) = \cos (u) \frac{d}{d x} (3 x)$

Replace all occurrences of $u$ with $3 x$ .

$f' (x) = \cos (3 x) \frac{d}{d x} (3 x)$

Differentiate.

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Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$f' (x) = \cos (3 x) (3 \frac{d}{d x} (x))$

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

$f' (x) = \cos (3 x) (3 \cdot 1)$

Simplify the expression.

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Multiply $3$ by $1$ .

$f' (x) = \cos (3 x) \cdot 3$

Move $3$ to the left of $\cos (3 x)$ .

$f' (x) = 3 \cos (3 x)$

Step 2

Find the second derivative.

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Since $3$ is constant with respect to $x$ , the derivative of $3 \cos (3 x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos (3 x)]$ .

$f'' (x) = 3 \frac{d}{d x} (\cos (3 x))$

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 $g (x) = 3 x$ .

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To apply the Chain Rule, set $u$ as $3 x$ .

$f'' (x) = 3 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (3 x))$

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

$f'' (x) = 3 (- \sin (u) \frac{d}{d x} (3 x))$

Replace all occurrences of $u$ with $3 x$ .

$f'' (x) = 3 (- \sin (3 x) \frac{d}{d x} (3 x))$

Differentiate.

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Multiply $- 1$ by $3$ .

$f'' (x) = - 3 (\sin (3 x) \frac{d}{d x} (3 x))$

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$f'' (x) = - 3 \sin (3 x) (3 \frac{d}{d x} (x))$

Multiply $3$ by $- 3$ .

$f'' (x) = - 9 \sin (3 x) \frac{d}{d x} x$

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

$f'' (x) = - 9 \sin (3 x) \cdot 1$

Multiply $- 9$ by $1$ .

$f'' (x) = - 9 \sin (3 x)$

Step 3

Find the third derivative.

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Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 \sin (3 x)$ with respect to $x$ is $- 9 \frac{d}{d x} [\sin (3 x)]$ .

$f''' (x) = - 9 \frac{d}{d x} \sin (3 x)$

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 $g (x) = 3 x$ .

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To apply the Chain Rule, set $u$ as $3 x$ .

$f''' (x) = - 9 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (3 x))$

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

$f''' (x) = - 9 (\cos (u) \frac{d}{d x} (3 x))$

Replace all occurrences of $u$ with $3 x$ .

$f''' (x) = - 9 (\cos (3 x) \frac{d}{d x} (3 x))$

Differentiate.

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Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$f''' (x) = - 9 \cos (3 x) (3 \frac{d}{d x} (x))$

Multiply $3$ by $- 9$ .

$f''' (x) = - 27 \cos (3 x) \frac{d}{d x} x$

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

$f''' (x) = - 27 \cos (3 x) \cdot 1$

Multiply $- 27$ by $1$ .

$f''' (x) = - 27 \cos (3 x)$

Step 4

The third derivative of $f (x)$ with respect to $x$ is $- 27 \cos (3 x)$ .

$- 27 \cos (3 x)$