Calculus Examples

$81 \cos (3 x)$

Find the first derivative.

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Since $81$ is constant with respect to $x$ , the derivative of $81 \cos (3 x)$ with respect to $x$ is $81 \frac{d}{d x} [\cos (3 x)]$ .

$f' (x) = 81 \frac{d}{d x} (\cos (3 x))$

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 $3 x$ .

$f' (x) = 81 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (3 x))$

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

$f' (x) = 81 (- \sin (u) \frac{d}{d x} (3 x))$

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

$f' (x) = 81 (- \sin (3 x) \frac{d}{d x} (3 x))$

Differentiate.

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Multiply $- 1$ by $81$ .

$f' (x) = - 81 (\sin (3 x) \frac{d}{d x} (3 x))$

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

$f' (x) = - 81 \sin (3 x) (3 \frac{d}{d x} (x))$

Multiply $3$ by $- 81$ .

$f' (x) = - 243 \sin (3 x) \frac{d}{d x} x$

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

$f' (x) = - 243 \sin (3 x) \cdot 1$

Multiply $- 243$ by $1$ .

$f' (x) = - 243 \sin (3 x)$

Find the second derivative.

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Since $- 243$ is constant with respect to $x$ , the derivative of $- 243 \sin (3 x)$ with respect to $x$ is $- 243 \frac{d}{d x} [\sin (3 x)]$ .

$f'' (x) = - 243 \frac{d}{d x} \sin (3 x)$

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 $3 x$ .

$f'' (x) = - 243 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (3 x))$

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

$f'' (x) = - 243 (\cos (u) \frac{d}{d x} (3 x))$

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

$f'' (x) = - 243 (\cos (3 x) \frac{d}{d x} (3 x))$

Differentiate.

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

$f'' (x) = - 243 \cos (3 x) (3 \frac{d}{d x} (x))$

Multiply $3$ by $- 243$ .

$f'' (x) = - 729 \cos (3 x) \frac{d}{d x} x$

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

$f'' (x) = - 729 \cos (3 x) \cdot 1$

Multiply $- 729$ by $1$ .

$f'' (x) = - 729 \cos (3 x)$

Find the third derivative.

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Since $- 729$ is constant with respect to $x$ , the derivative of $- 729 \cos (3 x)$ with respect to $x$ is $- 729 \frac{d}{d x} [\cos (3 x)]$ .

$f''' (x) = - 729 \frac{d}{d x} \cos (3 x)$

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 $3 x$ .

$f''' (x) = - 729 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (3 x))$

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

$f''' (x) = - 729 (- \sin (u) \frac{d}{d x} (3 x))$

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

$f''' (x) = - 729 (- \sin (3 x) \frac{d}{d x} (3 x))$

Differentiate.

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Multiply $- 1$ by $- 729$ .

$f''' (x) = 729 (\sin (3 x) \frac{d}{d x} (3 x))$

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

$f''' (x) = 729 \sin (3 x) (3 \frac{d}{d x} (x))$

Multiply $3$ by $729$ .

$f''' (x) = 2187 \sin (3 x) \frac{d}{d x} (x)$

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

$f''' (x) = 2187 \sin (3 x) \cdot 1$

Multiply $2187$ by $1$ .

$f''' (x) = 2187 \sin (3 x)$