Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $\frac{1}{2} \cdot \sin (π x - π) + \frac{1}{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2} \cdot \sin (π x - π)] + \frac{d}{d x} [\frac{1}{2}]$ .

$\frac{d}{d x} [\frac{1}{2} \cdot \sin (π x - π)] + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2

Evaluate $\frac{d}{d x} [\frac{1}{2} \cdot \sin (π x - π)]$ .

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Step 3.2.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} \cdot \sin (π x - π)$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\sin (π x - π)]$ .

$\frac{1}{2} \frac{d}{d x} [\sin (π x - π)] + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.2

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 $g (x) = π x - π$ .

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Step 3.2.2.1

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

$\frac{1}{2} (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [π x - π]) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.2.2

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

$\frac{1}{2} (\cos (u) \frac{d}{d x} [π x - π]) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.2.3

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

$\frac{1}{2} (\cos (π x - π) \frac{d}{d x} [π x - π]) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.3

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

$\frac{1}{2} (\cos (π x - π) (\frac{d}{d x} [π x] + \frac{d}{d x} [- π])) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.4

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

$\frac{1}{2} (\cos (π x - π) (π \frac{d}{d x} [x] + \frac{d}{d x} [- π])) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.5

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

$\frac{1}{2} (\cos (π x - π) (π \cdot 1 + \frac{d}{d x} [- π])) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.6

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

$\frac{1}{2} (\cos (π x - π) (π \cdot 1 + 0)) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.7

Multiply $π$ by $1$ .

$\frac{1}{2} (\cos (π x - π) (π + 0)) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.8

Add $π$ and $0$ .

$\frac{1}{2} (\cos (π x - π) π) + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.9

Combine $\cos (π x - π)$ and $\frac{1}{2}$ .

$\frac{\cos (π x - π)}{2} π + \frac{d}{d x} [\frac{1}{2}]$

Step 3.2.10

Combine $\frac{\cos (π x - π)}{2}$ and $π$ .

$\frac{\cos (π x - π) π}{2} + \frac{d}{d x} [\frac{1}{2}]$

Step 3.3

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2}$ with respect to $x$ is $0$ .

$\frac{\cos (π x - π) π}{2} + 0$

Step 3.4

Simplify.

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Step 3.4.1

Add $\frac{\cos (π x - π) π}{2}$ and $0$ .

$\frac{\cos (π x - π) π}{2}$

Step 3.4.2

Reorder terms.

$\frac{π \cos (π x - π)}{2}$

Step 4

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

$y' = \frac{π \cos (π x - π)}{2}$

Step 5

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

$\frac{d y}{d x} = \frac{π \cos (π x - π)}{2}$