Calculus Examples

$y = 8 e^{x} + x$ , $(0, 8)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 8$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of $8 e^{x} + x$ with respect to $x$ is $\frac{d}{d x} [8 e^{x}] + \frac{d}{d x} [x]$ .

$\frac{d}{d x} [8 e^{x}] + \frac{d}{d x} [x]$

Step 1.2

Evaluate $\frac{d}{d x} [8 e^{x}]$ .

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Step 1.2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 e^{x}$ with respect to $x$ is $8 \frac{d}{d x} [e^{x}]$ .

$8 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [x]$

Step 1.2.2

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

$8 e^{x} + \frac{d}{d x} [x]$

Step 1.3

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

$8 e^{x} + 1$

Step 1.4

Evaluate the derivative at $x = 0$ .

$8 e^{0} + 1$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Anything raised to $0$ is $1$ .

$8 \cdot 1 + 1$

Step 1.5.1.2

Multiply $8$ by $1$ .

$8 + 1$

Step 1.5.2

Add $8$ and $1$ .

$9$

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 $9$ and a given point $(0, 8)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (8) = 9 \cdot (x - (0))$

Step 2.2

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

$y - 8 = 9 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 8 = 9 x$

Step 2.3.2

Add $8$ to both sides of the equation.

$y = 9 x + 8$

Step 3