Calculus Examples

$y = \sin (x) + \cos (x)$ , $(0, 1)$

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

By the Sum Rule, the derivative of $\sin (x) + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [\cos (x)]$ .

$\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [\cos (x)]$

Step 1.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\cos (x) + \frac{d}{d x} [\cos (x)]$

Step 1.3

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\cos (x) - \sin (x)$

Step 1.4

Evaluate the derivative at $x = 0$ .

$\cos (0) - \sin (0)$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

The exact value of $\cos (0)$ is $1$ .

$1 - \sin (0)$

Step 1.5.1.2

The exact value of $\sin (0)$ is $0$ .

$1 - 0$

Step 1.5.1.3

Multiply $- 1$ by $0$ .

$1 + 0$

Step 1.5.2

Add $1$ and $0$ .

$1$

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

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

Step 2.2

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

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $1 \cdot (x + 0)$ .

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Step 2.3.1.1

Multiply $x + 0$ by $1$ .

$y - 1 = x + 0$

Step 2.3.1.2

Add $x$ and $0$ .

$y - 1 = x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = x + 1$

Step 3