Calculus Examples

$y = {(e^{2 x} - 4)}^{2}$ , $(0, 9)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 9$ 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 $f (x) = x^{2}$ and $g (x) = e^{2 x} - 4$ .

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Step 1.1.1

To apply the Chain Rule, set $u_{1}$ as $e^{2 x} - 4$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [e^{2 x} - 4]$

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 $n = 2$ .

$2 u_{1} \frac{d}{d x} [e^{2 x} - 4]$

Step 1.1.3

Replace all occurrences of $u_{1}$ with $e^{2 x} - 4$ .

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

Step 1.2

By the Sum Rule, the derivative of $e^{2 x} - 4$ with respect to $x$ is $\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- 4]$ .

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

Step 1.3

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

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Step 1.3.1

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

$2 (e^{2 x} - 4) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [2 x] + \frac{d}{d x} [- 4])$

Step 1.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$2 (e^{2 x} - 4) (e^{u_{2}} \frac{d}{d x} [2 x] + \frac{d}{d x} [- 4])$

Step 1.3.3

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

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

Step 1.4

Differentiate.

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Step 1.4.1

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

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

Step 1.4.2

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

$2 (e^{2 x} - 4) (e^{2 x} (2 \cdot 1) + \frac{d}{d x} [- 4])$

Step 1.4.3

Simplify the expression.

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Step 1.4.3.1

Multiply $2$ by $1$ .

$2 (e^{2 x} - 4) (e^{2 x} \cdot 2 + \frac{d}{d x} [- 4])$

Step 1.4.3.2

Move $2$ to the left of $e^{2 x}$ .

$2 (e^{2 x} - 4) (2 \cdot e^{2 x} + \frac{d}{d x} [- 4])$

Step 1.4.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$2 (e^{2 x} - 4) (2 e^{2 x} + 0)$

Step 1.4.5

Simplify the expression.

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Step 1.4.5.1

Add $2 e^{2 x}$ and $0$ .

$2 (e^{2 x} - 4) (2 e^{2 x})$

Step 1.4.5.2

Multiply $2$ by $2$ .

$4 (e^{2 x} - 4) e^{2 x}$

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

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

Step 1.5.2

Apply the distributive property.

$4 e^{2 x} e^{2 x} + 4 \cdot - 4 e^{2 x}$

Step 1.5.3

Combine terms.

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Step 1.5.3.1

Multiply $e^{2 x}$ by $e^{2 x}$ by adding the exponents.

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Step 1.5.3.1.1

Move $e^{2 x}$ .

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

Step 1.5.3.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 e^{2 x + 2 x} + 4 \cdot - 4 e^{2 x}$

Step 1.5.3.1.3

Add $2 x$ and $2 x$ .

$4 e^{4 x} + 4 \cdot - 4 e^{2 x}$

Step 1.5.3.2

Multiply $4$ by $- 4$ .

$4 e^{4 x} - 16 e^{2 x}$

Step 1.6

Evaluate the derivative at $x = 0$ .

$4 e^{4 (0)} - 16 e^{2 (0)}$

Step 1.7

Simplify.

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Step 1.7.1

Simplify each term.

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Step 1.7.1.1

Multiply $4$ by $0$ .

$4 e^{0} - 16 e^{2 (0)}$

Step 1.7.1.2

Anything raised to $0$ is $1$ .

$4 \cdot 1 - 16 e^{2 (0)}$

Step 1.7.1.3

Multiply $4$ by $1$ .

$4 - 16 e^{2 (0)}$

Step 1.7.1.4

Multiply $2$ by $0$ .

$4 - 16 e^{0}$

Step 1.7.1.5

Anything raised to $0$ is $1$ .

$4 - 16 \cdot 1$

Step 1.7.1.6

Multiply $- 16$ by $1$ .

$4 - 16$

Step 1.7.2

Subtract $16$ from $4$ .

$- 12$

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 $- 12$ and a given point $(0, 9)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (9) = - 12 \cdot (x - (0))$

Step 2.2

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

$y - 9 = - 12 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 9 = - 12 x$

Step 2.3.2

Add $9$ to both sides of the equation.

$y = - 12 x + 9$

Step 3