Calculus Examples

$y = {(x^{3} - 4 x)}^{8}$ at the point $(2, 0)$

Step 1

Find the first derivative and evaluate at $x = 2$ and $y = 0$ to find the slope of the tangent line.

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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 $f (x) = x^{8}$ and $g (x) = x^{3} - 4 x$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $x^{3} - 4 x$ .

$\frac{d}{d u} [u^{8}] \frac{d}{d x} [x^{3} - 4 x]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 8$ .

$8 u^{7} \frac{d}{d x} [x^{3} - 4 x]$

Step 1.1.3

Replace all occurrences of $u$ with $x^{3} - 4 x$ .

$8 {(x^{3} - 4 x)}^{7} \frac{d}{d x} [x^{3} - 4 x]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{3} - 4 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4 x]$ .

$8 {(x^{3} - 4 x)}^{7} (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4 x])$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$8 {(x^{3} - 4 x)}^{7} (3 x^{2} + \frac{d}{d x} [- 4 x])$

Step 1.2.3

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

$8 {(x^{3} - 4 x)}^{7} (3 x^{2} - 4 \frac{d}{d x} [x])$

Step 1.2.4

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

$8 {(x^{3} - 4 x)}^{7} (3 x^{2} - 4 \cdot 1)$

Step 1.2.5

Multiply $- 4$ by $1$ .

$8 {(x^{3} - 4 x)}^{7} (3 x^{2} - 4)$

Step 1.3

Evaluate the derivative at $x = 2$ .

$8 {({(2)}^{3} - 4 \cdot 2)}^{7} (3 {(2)}^{2} - 4)$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Raise $2$ to the power of $3$ .

$8 {(8 - 4 \cdot 2)}^{7} (3 {(2)}^{2} - 4)$

Step 1.4.1.2

Multiply $- 4$ by $2$ .

$8 {(8 - 8)}^{7} (3 {(2)}^{2} - 4)$

Step 1.4.2

Simplify the expression.

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Step 1.4.2.1

Subtract $8$ from $8$ .

$8 \cdot 0^{7} (3 {(2)}^{2} - 4)$

Step 1.4.2.2

Raising $0$ to any positive power yields $0$ .

$8 \cdot 0 (3 {(2)}^{2} - 4)$

Step 1.4.2.3

Multiply $8$ by $0$ .

$0 (3 {(2)}^{2} - 4)$

Step 1.4.3

Simplify each term.

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Step 1.4.3.1

Raise $2$ to the power of $2$ .

$0 (3 \cdot 4 - 4)$

Step 1.4.3.2

Multiply $3$ by $4$ .

$0 (12 - 4)$

Step 1.4.4

Simplify the expression.

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Step 1.4.4.1

Subtract $4$ from $12$ .

$0 \cdot 8$

Step 1.4.4.2

Multiply $0$ by $8$ .

$0$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $0$ and a given point $(2, 0)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = 0 \cdot (x - (2))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y + 0 = 0 \cdot (x - 2)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

$y = 0 \cdot (x - 2)$

Step 2.3.2

Multiply $0$ by $x - 2$ .

$y = 0$

Step 3