Calculus Examples

$y = 7 + \cot (x) - 2 \csc (x)$ , $(\frac{π}{2}, 5)$

Step 1

Find the first derivative and evaluate at $x = \frac{π}{2}$ and $y = 5$ 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 $7 + \cot (x) - 2 \csc (x)$ with respect to $x$ is $\frac{d}{d x} [7] + \frac{d}{d x} [\cot (x)] + \frac{d}{d x} [- 2 \csc (x)]$ .

$\frac{d}{d x} [7] + \frac{d}{d x} [\cot (x)] + \frac{d}{d x} [- 2 \csc (x)]$

Step 1.1.2

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\cot (x)] + \frac{d}{d x} [- 2 \csc (x)]$

Step 1.2

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

$0 - \csc^{2} (x) + \frac{d}{d x} [- 2 \csc (x)]$

Step 1.3

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

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Step 1.3.1

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

$0 - \csc^{2} (x) - 2 \frac{d}{d x} [\csc (x)]$

Step 1.3.2

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

$0 - \csc^{2} (x) - 2 (- \csc (x) \cot (x))$

Step 1.3.3

Multiply $- 1$ by $- 2$ .

$0 - \csc^{2} (x) + 2 \csc (x) \cot (x)$

Step 1.4

Simplify.

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Step 1.4.1

Subtract $\csc^{2} (x)$ from $0$ .

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

Step 1.4.2

Reorder terms.

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

Step 1.5

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

$2 \cot (\frac{π}{2}) \csc (\frac{π}{2}) - \csc^{2} (\frac{π}{2})$

Step 1.6

Simplify.

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Step 1.6.1

Simplify each term.

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Step 1.6.1.1

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

$2 \cdot 0 \csc (\frac{π}{2}) - \csc^{2} (\frac{π}{2})$

Step 1.6.1.2

Multiply $2$ by $0$ .

$0 \csc (\frac{π}{2}) - \csc^{2} (\frac{π}{2})$

Step 1.6.1.3

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

$0 \cdot 1 - \csc^{2} (\frac{π}{2})$

Step 1.6.1.4

Multiply $0$ by $1$ .

$0 - \csc^{2} (\frac{π}{2})$

Step 1.6.1.5

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

$0 - 1^{2}$

Step 1.6.1.6

One to any power is one.

$0 - 1 \cdot 1$

Step 1.6.1.7

Multiply $- 1$ by $1$ .

$0 - 1$

Step 1.6.2

Subtract $1$ from $0$ .

$- 1$

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 $(\frac{π}{2}, 5)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = - 1 \cdot (x - (\frac{π}{2}))$

Step 2.2

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

$y - 5 = - 1 \cdot (x - \frac{π}{2})$

Step 2.3

Solve for $y$ .

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Step 2.3.1

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

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Step 2.3.1.1

Rewrite.

$y - 5 = 0 + 0 - 1 \cdot (x - \frac{π}{2})$

Step 2.3.1.2

Simplify by adding zeros.

$y - 5 = - 1 \cdot (x - \frac{π}{2})$

Step 2.3.1.3

Apply the distributive property.

$y - 5 = - 1 x - 1 (- \frac{π}{2})$

Step 2.3.1.4

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

$y - 5 = - x - 1 (- \frac{π}{2})$

Step 2.3.1.5

Multiply $- 1 (- \frac{π}{2})$ .

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Step 2.3.1.5.1

Multiply $- 1$ by $- 1$ .

$y - 5 = - x + 1 \frac{π}{2}$

Step 2.3.1.5.2

Multiply $\frac{π}{2}$ by $1$ .

$y - 5 = - x + \frac{π}{2}$

Step 2.3.2

Add $5$ to both sides of the equation.

$y = - x + \frac{π}{2} + 5$

Step 2.3.3

Write in $y = m x + b$ form.

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Step 2.3.3.1

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = - x + \frac{π}{2} + 5 \cdot \frac{2}{2}$

Step 2.3.3.2

Combine $5$ and $\frac{2}{2}$ .

$y = - x + \frac{π}{2} + \frac{5 \cdot 2}{2}$

Step 2.3.3.3

Combine the numerators over the common denominator.

$y = - x + \frac{π + 5 \cdot 2}{2}$

Step 2.3.3.4

Multiply $5$ by $2$ .

$y = - x + \frac{π + 10}{2}$

Step 3