Calculus Examples

$y = \sin^{2} (x)$

Step 1

Find the first derivative.

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Step 1.1

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

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Step 1.1.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 1.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

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

Step 1.2

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

$2 \sin (x) \cos (x)$

Step 1.3

Simplify.

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Step 1.3.1

Reorder the factors of $2 \sin (x) \cos (x)$ .

$2 \cos (x) \sin (x)$

Step 1.3.2

Reorder $2 \cos (x)$ and $\sin (x)$ .

$\sin (x) (2 \cos (x))$

Step 1.3.3

Reorder $\sin (x)$ and $2$ .

$2 \cdot \sin (x) \cos (x)$

Step 1.3.4

Apply the sine double-angle identity.

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

Step 2

Find the second derivative.

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Step 2.1

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 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\cos (2 x) (2 \cdot 1)$

Step 2.2.3

Simplify the expression.

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Step 2.2.3.1

Multiply $2$ by $1$ .

$\cos (2 x) \cdot 2$

Step 2.2.3.2

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

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

Step 3

Find the third derivative.

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Step 3.1

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

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

Step 3.2

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 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 1$ by $2$ .

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

Step 3.3.2

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

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

Step 3.3.3

Multiply $2$ by $- 2$ .

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

Step 3.3.4

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

$- 4 \sin (2 x) \cdot 1$

Step 3.3.5

Multiply $- 4$ by $1$ .

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