Calculus Examples

$y = x^{2} - 2 x + 1$ , $(0, 1)$

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.

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Step 1.1.1

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

Step 1.2

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 1.2.1

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

$2 x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 1.2.2

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

$2 x - 2 \cdot 1 + \frac{d}{d x} [1]$

Step 1.2.3

Multiply $- 2$ by $1$ .

$2 x - 2 + \frac{d}{d x} [1]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 x - 2 + 0$

Step 1.3.2

Add $2 x - 2$ and $0$ .

$2 x - 2$

Step 1.4

Evaluate the derivative at $x = 0$ .

$2 (0) - 2$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $2$ by $0$ .

$0 - 2$

Step 1.5.2

Subtract $2$ from $0$ .

$- 2$

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 $- 2$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = - 2 \cdot (x - (0))$

Step 2.2

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

$y - 1 = - 2 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 1 = - 2 x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = - 2 x + 1$

Step 3

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