Calculus Examples

$y = \cot (x)$ ; $x = - \frac{π}{2}$

Step 1

Find the corresponding $y$ -value to $x = - \frac{π}{2}$ .

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Step 1.1

Substitute $- \frac{π}{2}$ in for $x$ .

$y = \cot (- \frac{π}{2})$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = \cot (- \frac{π}{2})$

Step 1.2.2

Simplify $\cot (- \frac{π}{2})$ .

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Step 1.2.2.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$y = \cot (\frac{3 π}{2})$

Step 1.2.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the fourth quadrant.

$y = - \cot (\frac{π}{2})$

Step 1.2.2.3

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$y = - 0$

Step 1.2.2.4

Multiply $- 1$ by $0$ .

$y = 0$

Step 2

Find the first derivative and evaluate at $x = - \frac{π}{2}$ and $y = 0$ to find the slope of the tangent line.

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Step 2.1

The derivative of $\cot (x)$ with respect to $x$ is $- \csc^{2} (x)$ .

$- \csc^{2} (x)$

Step 2.2

Evaluate the derivative at $x = - \frac{π}{2}$ .

$- \csc^{2} (- \frac{π}{2})$

Step 2.3

Simplify.

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Step 2.3.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \csc^{2} (\frac{3 π}{2})$

Step 2.3.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

$- {(- \csc (\frac{π}{2}))}^{2}$

Step 2.3.3

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$- {(- 1 \cdot 1)}^{2}$

Step 2.3.4

Multiply $- 1$ by $1$ .

$- {(- 1)}^{2}$

Step 2.3.5

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 2.3.5.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 2.3.5.1.1

Raise $- 1$ to the power of $1$ .

${(- 1)}^{1} {(- 1)}^{2}$

Step 2.3.5.1.2

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

${(- 1)}^{1 + 2}$

Step 2.3.5.2

Add $1$ and $2$ .

${(- 1)}^{3}$

Step 2.3.6

Raise $- 1$ to the power of $3$ .

$- 1$

Step 3

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 3.1

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

$y - (0) = - 1 \cdot (x - (- \frac{π}{2}))$

Step 3.2

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

$y + 0 = - 1 \cdot (x + \frac{π}{2})$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Add $y$ and $0$ .

$y = - 1 \cdot (x + \frac{π}{2})$

Step 3.3.2

Simplify $- 1 \cdot (x + \frac{π}{2})$ .

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Step 3.3.2.1

Apply the distributive property.

$y = - 1 x - 1 \frac{π}{2}$

Step 3.3.2.2

Rewrite negatives.

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Step 3.3.2.2.1

Rewrite $- 1 x$ as $- x$ .

$y = - x - 1 \frac{π}{2}$

Step 3.3.2.2.2

Rewrite $- 1 \frac{π}{2}$ as $- \frac{π}{2}$ .

$y = - x - \frac{π}{2}$

Step 4