Find the 3rd Derivative f(x)=sin(2x)

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

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 $g (x) = 2 x$ .

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

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

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

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

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

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

Differentiate.

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

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

Move $2$ to the left of $\cos (2 x)$ .

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

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

$f' (x) = 2 \cos (2 x) \cdot 1$

Multiply $2$ by $1$ .

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

Find the second derivative.

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

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

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 $g (x) = 2 x$ .

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

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

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

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

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

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

Differentiate.

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Multiply $- 1$ by $2$ .

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

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

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

Multiply $2$ by $- 2$ .

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

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

$f'' (x) = - 4 \sin (2 x) \cdot 1$

Multiply $- 4$ by $1$ .

$f'' (x) = - 4 \sin (2 x)$

Find the third derivative.

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

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

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 $g (x) = 2 x$ .

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

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

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

$f''' (x) = - 4 (\cos (u) \frac{d}{d x} (2 x))$

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

$f''' (x) = - 4 (\cos (2 x) \frac{d}{d x} (2 x))$

Differentiate.

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

$f''' (x) = - 4 \cos (2 x) (2 \frac{d}{d x} (x))$

Multiply $2$ by $- 4$ .

$f''' (x) = - 8 \cos (2 x) \frac{d}{d x} x$

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

$f''' (x) = - 8 \cos (2 x) \cdot 1$

Multiply $- 8$ by $1$ .

$f''' (x) = - 8 \cos (2 x)$

Find the fourth derivative.

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Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 \cos (2 x)$ with respect to $x$ is $- 8 \frac{d}{d x} [\cos (2 x)]$ .

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

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 $g (x) = 2 x$ .

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

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

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

$f^{4} (x) = - 8 (- \sin (u) \frac{d}{d x} (2 x))$

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

$f^{4} (x) = - 8 (- \sin (2 x) \frac{d}{d x} (2 x))$

Differentiate.

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Multiply $- 1$ by $- 8$ .

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

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

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

Multiply $2$ by $8$ .

$f^{4} (x) = 16 \sin (2 x) \frac{d}{d x} (x)$

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

$f^{4} (x) = 16 \sin (2 x) \cdot 1$

Multiply $16$ by $1$ .

$f^{4} (x) = 16 \sin (2 x)$

The 4th derivative of $f (x)$ with respect to $x$ is $16 \sin (2 x)$ .

$16 \sin (2 x)$

Find the 3rd Derivative f(x)=sin(2x)

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