Calculus Examples

$y = \tan (x)$ ; $x = - π$

Step 1

Find the corresponding $y$ -value to $x = - π$ .

Tap for more steps...

Step 1.1

Substitute $- π$ in for $x$ .

$y = \tan (- π)$

Step 1.2

Solve for $y$ .

Tap for more steps...

Step 1.2.1

Remove parentheses.

$y = \tan (- π)$

Step 1.2.2

Simplify $\tan (- π)$ .

Tap for more steps...

Step 1.2.2.1

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

$y = - \tan (0)$

Step 1.2.2.2

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

$y = - 0$

Step 1.2.2.3

Multiply $- 1$ by $0$ .

$y = 0$

Step 2

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

Tap for more steps...

Step 2.1

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$\sec^{2} (x)$

Step 2.2

Evaluate the derivative at $x = - π$ .

$\sec^{2} (- π)$

Step 2.3

Simplify.

Tap for more steps...

Step 2.3.1

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

${(- \sec (0))}^{2}$

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.3.4

Raise $- 1$ to the power of $2$ .

$1$

Step 3

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

Tap for more steps...

Step 3.1

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

$y - (0) = 1 \cdot (x - (- π))$

Step 3.2

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

$y + 0 = 1 \cdot (x + π)$

Step 3.3

Solve for $y$ .

Tap for more steps...

Step 3.3.1

Add $y$ and $0$ .

$y = 1 \cdot (x + π)$

Step 3.3.2

Multiply $x + π$ by $1$ .

$y = x + π$

Step 4