Calculus Examples

$\sin (y) = x$ ; $(0, π)$

Step 1

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

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Step 1.1

Differentiate both sides of the equation.

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

Step 1.2

Differentiate the left side of the equation.

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Step 1.2.1

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 $g (x) = y$ .

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.1.3

Replace all occurrences of $u$ with $y$ .

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

Step 1.2.2

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

$\cos (y) y'$

Step 1.3

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

$1$

Step 1.4

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

$\cos (y) y' = 1$

Step 1.5

Divide each term in $\cos (y) y' = 1$ by $\cos (y)$ and simplify.

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Step 1.5.1

Divide each term in $\cos (y) y' = 1$ by $\cos (y)$ .

$\frac{\cos (y) y'}{\cos (y)} = \frac{1}{\cos (y)}$

Step 1.5.2

Simplify the left side.

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Step 1.5.2.1

Cancel the common factor of $\cos (y)$ .

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Step 1.5.2.1.1

Cancel the common factor.

$\frac{\cos (y) y'}{\cos (y)} = \frac{1}{\cos (y)}$

Step 1.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{1}{\cos (y)}$

Step 1.5.3

Simplify the right side.

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Step 1.5.3.1

Convert from $\frac{1}{\cos (y)}$ to $\sec (y)$ .

$y' = \sec (y)$

Step 1.6

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

$\frac{d y}{d x} = \sec (y)$

Step 1.7

Evaluate at $x = 0$ and $y = π$ .

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Step 1.7.1

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

$\sec (y)$

Step 1.7.2

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

$\sec (π)$

Step 1.7.3

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.

$- \sec (0)$

Step 1.7.4

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

$- 1 \cdot 1$

Step 1.7.5

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

$y - (π) = - 1 \cdot (x - (0))$

Step 2.2

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

$y - π = - 1 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $- 1 \cdot (x + 0)$ .

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Step 2.3.1.1

Add $x$ and $0$ .

$y - π = - 1 \cdot x$

Step 2.3.1.2

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

$y - π = - x$

Step 2.3.2

Add $π$ to both sides of the equation.

$y = - x + π$

Step 3