Calculus Examples

$y = e^{3 x}$ , $(0, 1)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 1$ 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 $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 1.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x]$

Step 1.1.2

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

$e^{u} \frac{d}{d x} [3 x]$

Step 1.1.3

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

$e^{3 x} \frac{d}{d x} [3 x]$

Step 1.2

Differentiate.

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Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$e^{3 x} (3 \frac{d}{d x} [x])$

Step 1.2.2

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

$e^{3 x} (3 \cdot 1)$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $3$ by $1$ .

$e^{3 x} \cdot 3$

Step 1.2.3.2

Move $3$ to the left of $e^{3 x}$ .

$3 e^{3 x}$

Step 1.3

Evaluate the derivative at $x = 0$ .

$3 e^{3 (0)}$

Step 1.4

Simplify.

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Step 1.4.1

Multiply $3$ by $0$ .

$3 e^{0}$

Step 1.4.2

Anything raised to $0$ is $1$ .

$3 \cdot 1$

Step 1.4.3

Multiply $3$ by $1$ .

$3$

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 $3$ and a given point $(0, 1)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 3 \cdot (x - (0))$

Step 2.2

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

$y - 1 = 3 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 1 = 3 x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = 3 x + 1$

Step 3