Calculus Examples

$f (x) = x^{3} {(3 - x)}^{4}$ ; $x = 2$

Step 1

Find the corresponding $y$ -value to $x = 2$ .

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Step 1.1

Substitute $2$ in for $x$ .

$y = {(2)}^{3} {(3 - (2))}^{4}$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = 2^{3} {(3 - (2))}^{4}$

Step 1.2.2

Remove parentheses.

$y = {(2)}^{3} {(3 - (2))}^{4}$

Step 1.2.3

Simplify ${(2)}^{3} {(3 - (2))}^{4}$ .

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Step 1.2.3.1

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

$y = 8 {(3 - (2))}^{4}$

Step 1.2.3.2

Multiply $- 1$ by $2$ .

$y = 8 {(3 - 2)}^{4}$

Step 1.2.3.3

Subtract $2$ from $3$ .

$y = 8 \cdot 1^{4}$

Step 1.2.3.4

One to any power is one.

$y = 8 \cdot 1$

Step 1.2.3.5

Multiply $8$ by $1$ .

$y = 8$

Step 2

Find the first derivative and evaluate at $x = 2$ and $y = 8$ to find the slope of the tangent line.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{3}$ and $g (x) = {(3 - x)}^{4}$ .

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = 3 - x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $3 - x$ .

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

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

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

Step 2.2.3

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

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

$x^{3} (4 {(3 - x)}^{3} (0 + \frac{d}{d x} [- x])) + {(3 - x)}^{4} \frac{d}{d x} [x^{3}]$

Step 2.3.3

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 2.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.3.5

Multiply $- 1$ by $4$ .

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

Step 2.3.6

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

$x^{3} (- 4 {(3 - x)}^{3} \cdot 1) + {(3 - x)}^{4} \frac{d}{d x} [x^{3}]$

Step 2.3.7

Multiply $- 4$ by $1$ .

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

Step 2.3.8

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

$x^{3} (- 4 {(3 - x)}^{3}) + {(3 - x)}^{4} (3 x^{2})$

Step 2.3.9

Move $3$ to the left of ${(3 - x)}^{4}$ .

$x^{3} (- 4 {(3 - x)}^{3}) + 3 \cdot {(3 - x)}^{4} x^{2}$

Step 2.4

Simplify.

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Step 2.4.1

Factor $x^{2} {(3 - x)}^{3}$ out of $x^{3} (- 4 {(3 - x)}^{3}) + 3 {(3 - x)}^{4} x^{2}$ .

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Step 2.4.1.1

Factor $x^{2} {(3 - x)}^{3}$ out of $x^{3} (- 4 {(3 - x)}^{3})$ .

$x^{2} {(3 - x)}^{3} (x \cdot (- 4)) + 3 {(3 - x)}^{4} x^{2}$

Step 2.4.1.2

Factor $x^{2} {(3 - x)}^{3}$ out of $3 {(3 - x)}^{4} x^{2}$ .

$x^{2} {(3 - x)}^{3} (x \cdot (- 4)) + x^{2} {(3 - x)}^{3} (3 (3 - x))$

Step 2.4.1.3

Factor $x^{2} {(3 - x)}^{3}$ out of $x^{2} {(3 - x)}^{3} (x \cdot (- 4)) + x^{2} {(3 - x)}^{3} (3 (3 - x))$ .

$x^{2} {(3 - x)}^{3} (x \cdot (- 4) + 3 (3 - x))$

Step 2.4.2

Move $- 4$ to the left of $x$ .

$x^{2} {(3 - x)}^{3} (- 4 x + 3 (3 - x))$

Step 2.5

Evaluate the derivative at $x = 2$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Simplify the expression.

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Step 2.6.1.1

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

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

Step 2.6.1.2

Multiply $- 1$ by $2$ .

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

Step 2.6.1.3

Subtract $2$ from $3$ .

$4 \cdot 1^{3} (- 4 \cdot 2 + 3 (3 - (2)))$

Step 2.6.1.4

One to any power is one.

$4 \cdot 1 (- 4 \cdot 2 + 3 (3 - (2)))$

Step 2.6.1.5

Multiply $4$ by $1$ .

$4 (- 4 \cdot 2 + 3 (3 - (2)))$

Step 2.6.2

Simplify each term.

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Step 2.6.2.1

Multiply $- 4$ by $2$ .

$4 (- 8 + 3 (3 - (2)))$

Step 2.6.2.2

Multiply $- 1$ by $2$ .

$4 (- 8 + 3 (3 - 2))$

Step 2.6.2.3

Subtract $2$ from $3$ .

$4 (- 8 + 3 \cdot 1)$

Step 2.6.2.4

Multiply $3$ by $1$ .

$4 (- 8 + 3)$

Step 2.6.3

Simplify the expression.

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Step 2.6.3.1

Add $- 8$ and $3$ .

$4 \cdot - 5$

Step 2.6.3.2

Multiply $4$ by $- 5$ .

$- 20$

Step 3

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

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Step 3.1

Use the slope $- 20$ and a given point $(2, 8)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (8) = - 20 \cdot (x - (2))$

Step 3.2

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

$y - 8 = - 20 \cdot (x - 2)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $- 20 \cdot (x - 2)$ .

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Step 3.3.1.1

Rewrite.

$y - 8 = 0 + 0 - 20 \cdot (x - 2)$

Step 3.3.1.2

Simplify by adding zeros.

$y - 8 = - 20 \cdot (x - 2)$

Step 3.3.1.3

Apply the distributive property.

$y - 8 = - 20 x - 20 \cdot - 2$

Step 3.3.1.4

Multiply $- 20$ by $- 2$ .

$y - 8 = - 20 x + 40$

Step 3.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 3.3.2.1

Add $8$ to both sides of the equation.

$y = - 20 x + 40 + 8$

Step 3.3.2.2

Add $40$ and $8$ .

$y = - 20 x + 48$

Step 4