Calculus Examples

$f (x, y) = \sin^{2} (x - 3 y)$

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 $g (x) = \sin (x - 3 y)$ .

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Step 1.1.1

To apply the Chain Rule, set $u_{1}$ as $\sin (x - 3 y)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\sin (x - 3 y)]$

Step 1.1.2

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

$2 u_{1} \frac{d}{d x} [\sin (x - 3 y)]$

Step 1.1.3

Replace all occurrences of $u_{1}$ with $\sin (x - 3 y)$ .

$2 \sin (x - 3 y) \frac{d}{d x} [\sin (x - 3 y)]$

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 $g (x) = x - 3 y$ .

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Step 1.2.1

To apply the Chain Rule, set $u_{2}$ as $x - 3 y$ .

$2 \sin (x - 3 y) (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [x - 3 y])$

Step 1.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$2 \sin (x - 3 y) (\cos (u_{2}) \frac{d}{d x} [x - 3 y])$

Step 1.2.3

Replace all occurrences of $u_{2}$ with $x - 3 y$ .

$2 \sin (x - 3 y) (\cos (x - 3 y) \frac{d}{d x} [x - 3 y])$

Step 1.3

Differentiate.

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Step 1.3.1

By the Sum Rule, the derivative of $x - 3 y$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3 y]$ .

$2 \sin (x - 3 y) \cos (x - 3 y) (\frac{d}{d x} [x] + \frac{d}{d x} [- 3 y])$

Step 1.3.2

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

$2 \sin (x - 3 y) \cos (x - 3 y) (1 + \frac{d}{d x} [- 3 y])$

Step 1.3.3

Since $- 3 y$ is constant with respect to $x$ , the derivative of $- 3 y$ with respect to $x$ is $0$ .

$2 \sin (x - 3 y) \cos (x - 3 y) (1 + 0)$

Step 1.3.4

Simplify the expression.

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Step 1.3.4.1

Add $1$ and $0$ .

$2 \sin (x - 3 y) \cos (x - 3 y) \cdot 1$

Step 1.3.4.2

Multiply $2$ by $1$ .

$2 \sin (x - 3 y) \cos (x - 3 y)$

Step 1.4

Simplify.

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Step 1.4.1

Reorder the factors of $2 \sin (x - 3 y) \cos (x - 3 y)$ .

$2 \cos (x - 3 y) \sin (x - 3 y)$

Step 1.4.2

Reorder $2 \cos (x - 3 y)$ and $\sin (x - 3 y)$ .

$\sin (x - 3 y) (2 \cos (x - 3 y))$

Step 1.4.3

Reorder $\sin (x - 3 y)$ and $2$ .

$2 \cdot \sin (x - 3 y) \cos (x - 3 y)$

Step 1.4.4

Apply the sine double-angle identity.

$\sin (2 (x - 3 y))$

Step 1.4.5

Apply the distributive property.

$\sin (2 x + 2 (- 3 y))$

Step 1.4.6

Multiply $- 3$ by $2$ .

$f' (x) = \sin (2 x - 6 y)$

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 $g (x) = 2 x - 6 y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $2 x - 6 y$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x - 6 y]$

Step 2.1.2

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

$\cos (u) \frac{d}{d x} [2 x - 6 y]$

Step 2.1.3

Replace all occurrences of $u$ with $2 x - 6 y$ .

$\cos (2 x - 6 y) \frac{d}{d x} [2 x - 6 y]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $2 x - 6 y$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6 y]$ .

$\cos (2 x - 6 y) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6 y])$

Step 2.2.2

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

$\cos (2 x - 6 y) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 6 y])$

Step 2.2.3

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

$\cos (2 x - 6 y) (2 \cdot 1 + \frac{d}{d x} [- 6 y])$

Step 2.2.4

Multiply $2$ by $1$ .

$\cos (2 x - 6 y) (2 + \frac{d}{d x} [- 6 y])$

Step 2.2.5

Since $- 6 y$ is constant with respect to $x$ , the derivative of $- 6 y$ with respect to $x$ is $0$ .

$\cos (2 x - 6 y) (2 + 0)$

Step 2.2.6

Simplify the expression.

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Step 2.2.6.1

Add $2$ and $0$ .

$\cos (2 x - 6 y) \cdot 2$

Step 2.2.6.2

Move $2$ to the left of $\cos (2 x - 6 y)$ .

$f'' (x) = 2 \cos (2 x - 6 y)$

Step 3

Find the third derivative.

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Step 3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \cos (2 x - 6 y)$ with respect to $x$ is $2 \frac{d}{d x} [\cos (2 x - 6 y)]$ .

$2 \frac{d}{d x} [\cos (2 x - 6 y)]$

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 $g (x) = 2 x - 6 y$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $2 x - 6 y$ .

$2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x - 6 y])$

Step 3.2.2

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

$2 (- \sin (u) \frac{d}{d x} [2 x - 6 y])$

Step 3.2.3

Replace all occurrences of $u$ with $2 x - 6 y$ .

$2 (- \sin (2 x - 6 y) \frac{d}{d x} [2 x - 6 y])$

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 1$ by $2$ .

$- 2 (\sin (2 x - 6 y) \frac{d}{d x} [2 x - 6 y])$

Step 3.3.2

By the Sum Rule, the derivative of $2 x - 6 y$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6 y]$ .

$- 2 \sin (2 x - 6 y) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6 y])$

Step 3.3.3

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

$- 2 \sin (2 x - 6 y) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 6 y])$

Step 3.3.4

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

$- 2 \sin (2 x - 6 y) (2 \cdot 1 + \frac{d}{d x} [- 6 y])$

Step 3.3.5

Multiply $2$ by $1$ .

$- 2 \sin (2 x - 6 y) (2 + \frac{d}{d x} [- 6 y])$

Step 3.3.6

Since $- 6 y$ is constant with respect to $x$ , the derivative of $- 6 y$ with respect to $x$ is $0$ .

$- 2 \sin (2 x - 6 y) (2 + 0)$

Step 3.3.7

Simplify the expression.

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Step 3.3.7.1

Add $2$ and $0$ .

$- 2 \sin (2 x - 6 y) \cdot 2$

Step 3.3.7.2

Multiply $2$ by $- 2$ .

$f''' (x) = - 4 \sin (2 x - 6 y)$