Calculus Examples

$y = 3 \sin (π x - y)$ , $(1, 0)$

Step 1

Find the first derivative and evaluate at $x = 1$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (3 \sin (π x - y))$

Step 1.2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 1.3

Differentiate the right side of the equation.

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Step 1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \sin (π x - y)$ with respect to $x$ is $3 \frac{d}{d x} [\sin (π x - y)]$ .

$3 \frac{d}{d x} [\sin (π x - y)]$

Step 1.3.2

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 $g (x) = π x - y$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $π x - y$ .

$3 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [π x - y])$

Step 1.3.2.2

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

$3 (\cos (u) \frac{d}{d x} [π x - y])$

Step 1.3.2.3

Replace all occurrences of $u$ with $π x - y$ .

$3 (\cos (π x - y) \frac{d}{d x} [π x - y])$

Step 1.3.3

Differentiate.

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Step 1.3.3.1

By the Sum Rule, the derivative of $π x - y$ with respect to $x$ is $\frac{d}{d x} [π x] + \frac{d}{d x} [- y]$ .

$3 \cos (π x - y) (\frac{d}{d x} [π x] + \frac{d}{d x} [- y])$

Step 1.3.3.2

Since $π$ is constant with respect to $x$ , the derivative of $π x$ with respect to $x$ is $π \frac{d}{d x} [x]$ .

$3 \cos (π x - y) (π \frac{d}{d x} [x] + \frac{d}{d x} [- y])$

Step 1.3.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$3 \cos (π x - y) (π \cdot 1 + \frac{d}{d x} [- y])$

Step 1.3.3.4

Multiply $π$ by $1$ .

$3 \cos (π x - y) (π + \frac{d}{d x} [- y])$

Step 1.3.3.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- y$ with respect to $x$ is $- \frac{d}{d x} [y]$ .

$3 \cos (π x - y) (π - \frac{d}{d x} [y])$

Step 1.3.4

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$3 \cos (π x - y) (π - y')$

Step 1.4

Reform the equation by setting the left side equal to the right side.

$y' = 3 \cos (π x - y) (π - y')$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

Simplify the right side.

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Step 1.5.1.1

Simplify $3 \cos (π x - y) (π - y')$ .

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Step 1.5.1.1.1

Apply the distributive property.

$y' = 3 \cos (π x - y) π + 3 \cos (π x - y) (- y')$

Step 1.5.1.1.2

Simplify the expression.

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Step 1.5.1.1.2.1

Multiply $- 1$ by $3$ .

$y' = 3 \cos (π x - y) π - 3 \cos (π x - y) y'$

Step 1.5.1.1.2.2

Reorder factors in $3 \cos (π x - y) π - 3 \cos (π x - y) y'$ .

$y' = 3 π \cos (π x - y) - 3 y' \cos (π x - y)$

Step 1.5.2

Add $3 y' \cos (π x - y)$ to both sides of the equation.

$y' + 3 y' \cos (π x - y) = 3 π \cos (π x - y)$

Step 1.5.3

Factor $y'$ out of $y' + 3 y' \cos (π x - y)$ .

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Step 1.5.3.1

Factor $y'$ out of ${y'}^{1}$ .

$y' \cdot 1 + 3 y' \cos (π x - y) = 3 π \cos (π x - y)$

Step 1.5.3.2

Factor $y'$ out of $3 y' \cos (π x - y)$ .

$y' \cdot 1 + y' (3 \cos (π x - y)) = 3 π \cos (π x - y)$

Step 1.5.3.3

Factor $y'$ out of $y' \cdot 1 + y' (3 \cos (π x - y))$ .

$y' (1 + 3 \cos (π x - y)) = 3 π \cos (π x - y)$

Step 1.5.4

Divide each term in $y' (1 + 3 \cos (π x - y)) = 3 π \cos (π x - y)$ by $1 + 3 \cos (π x - y)$ and simplify.

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Step 1.5.4.1

Divide each term in $y' (1 + 3 \cos (π x - y)) = 3 π \cos (π x - y)$ by $1 + 3 \cos (π x - y)$ .

$\frac{y' (1 + 3 \cos (π x - y))}{1 + 3 \cos (π x - y)} = \frac{3 π \cos (π x - y)}{1 + 3 \cos (π x - y)}$

Step 1.5.4.2

Simplify the left side.

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Step 1.5.4.2.1

Cancel the common factor of $1 + 3 \cos (π x - y)$ .

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Step 1.5.4.2.1.1

Cancel the common factor.

$\frac{y' (1 + 3 \cos (π x - y))}{1 + 3 \cos (π x - y)} = \frac{3 π \cos (π x - y)}{1 + 3 \cos (π x - y)}$

Step 1.5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{3 π \cos (π x - y)}{1 + 3 \cos (π x - y)}$

Step 1.6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{3 π \cos (π x - y)}{1 + 3 \cos (π x - y)}$

Step 1.7

Evaluate at $x = 1$ and $y = 0$ .

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Step 1.7.1

Replace the variable $x$ with $1$ in the expression.

$\frac{3 π \cos (π (1) - y)}{1 + 3 \cos (π (1) - y)}$

Step 1.7.2

Replace the variable $y$ with $0$ in the expression.

$\frac{3 π \cos (π (1) - (0))}{1 + 3 \cos (π (1) - (0))}$

Step 1.7.3

Simplify the numerator.

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Step 1.7.3.1

Combine exponents.

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Step 1.7.3.1.1

Multiply $π$ by $1$ .

$\frac{3 π \cos (π - (0))}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.1.2

Multiply $- 1$ by $0$ .

$\frac{3 π \cos (π + 0)}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.2

Add $π$ and $0$ .

$\frac{3 π \cos (π)}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.3

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.

$\frac{3 π (- \cos (0))}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.4

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

$\frac{3 π (- 1 \cdot 1)}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.5

Multiply $- 1$ by $1$ .

$\frac{3 π \cdot - 1}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.3.6

Multiply $- 1$ by $3$ .

$\frac{- 3 π}{1 + 3 \cos (π \cdot 1 - (0))}$

Step 1.7.4

Simplify the denominator.

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Step 1.7.4.1

Multiply $π$ by $1$ .

$\frac{- 3 π}{1 + 3 \cos (π - (0))}$

Step 1.7.4.2

Subtract $0$ from $π$ .

$\frac{- 3 π}{1 + 3 \cos (π)}$

Step 1.7.4.3

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.

$\frac{- 3 π}{1 + 3 (- \cos (0))}$

Step 1.7.4.4

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

$\frac{- 3 π}{1 + 3 (- 1 \cdot 1)}$

Step 1.7.4.5

Multiply $3 (- 1 \cdot 1)$ .

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Step 1.7.4.5.1

Multiply $- 1$ by $1$ .

$\frac{- 3 π}{1 + 3 \cdot - 1}$

Step 1.7.4.5.2

Multiply $3$ by $- 1$ .

$\frac{- 3 π}{1 - 3}$

Step 1.7.4.6

Subtract $3$ from $1$ .

$\frac{- 3 π}{- 2}$

Step 1.7.5

Dividing two negative values results in a positive value.

$\frac{3 π}{2}$

Step 2

The normal line is perpendicular to the tangent line. Take the negative reciprocal of the slope of the tangent line to find the slope of the normal line.

$- \frac{2}{3 π}$

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

$y - (0) = - \frac{2}{3 π} \cdot (x - (1))$

Step 3.2

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

$y + 0 = - \frac{2}{3 π} \cdot (x - 1)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Add $y$ and $0$ .

$y = - \frac{2}{3 π} \cdot (x - 1)$

Step 3.3.2

Simplify $- \frac{2}{3 π} \cdot (x - 1)$ .

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Step 3.3.2.1

Apply the distributive property.

$y = - \frac{2}{3 π} x - \frac{2}{3 π} \cdot - 1$

Step 3.3.2.2

Combine $x$ and $\frac{2}{3 π}$ .

$y = - \frac{x \cdot 2}{3 π} - \frac{2}{3 π} \cdot - 1$

Step 3.3.2.3

Multiply $- \frac{2}{3 π} \cdot - 1$ .

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Step 3.3.2.3.1

Multiply $- 1$ by $- 1$ .

$y = - \frac{x \cdot 2}{3 π} + 1 \frac{2}{3 π}$

Step 3.3.2.3.2

Multiply $\frac{2}{3 π}$ by $1$ .

$y = - \frac{x \cdot 2}{3 π} + \frac{2}{3 π}$

Step 3.3.2.4

Move $2$ to the left of $x$ .

$y = - \frac{2 x}{3 π} + \frac{2}{3 π}$

Step 3.3.3

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

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Step 3.3.3.1

Reorder terms.

$y = - (\frac{2}{3 π} x) + \frac{2}{3 π}$

Step 3.3.3.2

Remove parentheses.

$y = - \frac{2}{3 π} x + \frac{2}{3 π}$

Step 4