Calculus Examples

$y = t \sin (π t) + \frac{1}{π} \cdot \cos (π t) - \frac{1}{π}$

Step 1

Differentiate both sides of the equation.

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

Step 2

The derivative of $y$ with respect to $t$ 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 $t \sin (π t) + \frac{1}{π} \cdot \cos (π t) - \frac{1}{π}$ with respect to $t$ is $\frac{d}{d t} [t \sin (π t)] + \frac{d}{d t} [\frac{1}{π} \cdot \cos (π t)] + \frac{d}{d t} [- \frac{1}{π}]$ .

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

Step 3.2

Evaluate $\frac{d}{d t} [t \sin (π t)]$ .

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Step 3.2.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = \sin (π t)$ .

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

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \sin (t)$ and $g (t) = π t$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $π t$ .

$t (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d t} [π t]) + \sin (π t) \frac{d}{d t} [t] + \frac{d}{d t} [\frac{1}{π} \cdot \cos (π t)] + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.2.2.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$t (\cos (u_{1}) \frac{d}{d t} [π t]) + \sin (π t) \frac{d}{d t} [t] + \frac{d}{d t} [\frac{1}{π} \cdot \cos (π t)] + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $π t$ .

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.2.6

Multiply $π$ by $1$ .

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

Step 3.2.7

Multiply $\sin (π t)$ by $1$ .

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

Step 3.3

Evaluate $\frac{d}{d t} [\frac{1}{π} \cdot \cos (π t)]$ .

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Step 3.3.1

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

$t (\cos (π t) π) + \sin (π t) + \frac{1}{π} \frac{d}{d t} [\cos (π t)] + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = π t$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $π t$ .

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

Step 3.3.2.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 3.3.2.3

Replace all occurrences of $u_{2}$ with $π t$ .

$t (\cos (π t) π) + \sin (π t) + \frac{1}{π} (- \sin (π t) \frac{d}{d t} [π t]) + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.3

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

$t (\cos (π t) π) + \sin (π t) + \frac{1}{π} (- \sin (π t) (π \frac{d}{d t} [t])) + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.4

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

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

Step 3.3.5

Multiply $π$ by $1$ .

$t (\cos (π t) π) + \sin (π t) + \frac{1}{π} (- \sin (π t) π) + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.6

Combine $\frac{1}{π}$ and $\sin (π t)$ .

$t (\cos (π t) π) + \sin (π t) - \frac{\sin (π t)}{π} π + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.7

Combine $π$ and $\frac{\sin (π t)}{π}$ .

$t (\cos (π t) π) + \sin (π t) - \frac{π \sin (π t)}{π} + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.8

Cancel the common factor of $π$ .

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Step 3.3.8.1

Cancel the common factor.

$t (\cos (π t) π) + \sin (π t) - \frac{π \sin (π t)}{π} + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.3.8.2

Divide $\sin (π t)$ by $1$ .

$t (\cos (π t) π) + \sin (π t) - \sin (π t) + \frac{d}{d t} [- \frac{1}{π}]$

Step 3.4

Since $- \frac{1}{π}$ is constant with respect to $t$ , the derivative of $- \frac{1}{π}$ with respect to $t$ is $0$ .

$t (\cos (π t) π) + \sin (π t) - \sin (π t) + 0$

Step 3.5

Simplify.

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Step 3.5.1

Combine terms.

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Step 3.5.1.1

Subtract $\sin (π t)$ from $\sin (π t)$ .

$t (\cos (π t) π) + 0 + 0$

Step 3.5.1.2

Add $t (\cos (π t) π)$ and $0$ .

$t (\cos (π t) π) + 0$

Step 3.5.1.3

Add $t (\cos (π t) π)$ and $0$ .

$t (\cos (π t) π)$

Step 3.5.2

Reorder the factors of $t (\cos (π t) π)$ .

$π t \cos (π t)$

Step 4

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

$y' = π t \cos (π t)$

Step 5

Replace $y'$ with $\frac{d y}{d t}$ .

$\frac{d y}{d t} = π t \cos (π t)$