Calculus Examples

$y = \sin (\sin (x))$ , $(3 π, 0)$

Step 1

Find the first derivative and evaluate at $x = 3 π$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = \sin (x)$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.1.2

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

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

Step 1.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

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

Step 1.2

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

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

Step 1.3

Reorder the factors of $\cos (\sin (x)) \cos (x)$ .

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

Step 1.4

Evaluate the derivative at $x = 3 π$ .

$\cos (3 π) \cos (\sin (3 π))$

Step 1.5

Simplify.

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Step 1.5.1

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

$\cos (π) \cos (\sin (3 π))$

Step 1.5.2

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

$- \cos (0) \cos (\sin (3 π))$

Step 1.5.3

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

$- 1 \cdot 1 \cos (\sin (3 π))$

Step 1.5.4

Multiply $- 1$ by $1$ .

$- 1 \cos (\sin (3 π))$

Step 1.5.5

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

$- 1 \cos (\sin (π))$

Step 1.5.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- 1 \cos (\sin (0))$

Step 1.5.7

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

$- 1 \cos (0)$

Step 1.5.8

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

$- 1 \cdot 1$

Step 1.5.9

Multiply $- 1$ by $1$ .

$- 1$

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 $(3 π, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - 1 \cdot (x - (3 π))$

Step 2.2

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

$y + 0 = - 1 \cdot (x - 3 π)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

$y = - 1 \cdot (x - 3 π)$

Step 2.3.2

Simplify $- 1 \cdot (x - 3 π)$ .

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Step 2.3.2.1

Apply the distributive property.

$y = - 1 x - 1 (- 3 π)$

Step 2.3.2.2

Simplify the expression.

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Step 2.3.2.2.1

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

$y = - x - 1 (- 3 π)$

Step 2.3.2.2.2

Multiply $- 3$ by $- 1$ .

$y = - x + 3 π$

Step 3