Calculus Examples

$P (t) = 4 e^{\frac{1}{2} t} - t + 1$

Step 1

Find the first derivative.

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Step 1.1

By the Sum Rule, the derivative of $4 e^{\frac{1}{2} t} - t + 1$ with respect to $t$ is $\frac{d}{d t} [4 e^{\frac{1}{2} t}] + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [4 e^{\frac{1}{2} t}] + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2

Evaluate $\frac{d}{d t} [4 e^{\frac{1}{2} t}]$ .

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Step 1.2.1

Combine $\frac{1}{2}$ and $t$ .

$\frac{d}{d t} [4 e^{\frac{t}{2}}] + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.2

Since $4$ is constant with respect to $t$ , the derivative of $4 e^{\frac{t}{2}}$ with respect to $t$ is $4 \frac{d}{d t} [e^{\frac{t}{2}}]$ .

$4 \frac{d}{d t} [e^{\frac{t}{2}}] + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = \frac{t}{2}$ .

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Step 1.2.3.1

To apply the Chain Rule, set $u$ as $\frac{t}{2}$ .

$4 (\frac{d}{d u} [e^{u}] \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$4 (e^{u} \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.3.3

Replace all occurrences of $u$ with $\frac{t}{2}$ .

$4 (e^{\frac{t}{2}} \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.4

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{t}{2}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [t]$ .

$4 (e^{\frac{t}{2}} (\frac{1}{2} \frac{d}{d t} [t])) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.5

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

$4 (e^{\frac{t}{2}} (\frac{1}{2} \cdot 1)) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.6

Multiply $\frac{1}{2}$ by $1$ .

$4 (e^{\frac{t}{2}} \frac{1}{2}) + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.7

Combine $e^{\frac{t}{2}}$ and $\frac{1}{2}$ .

$4 \frac{e^{\frac{t}{2}}}{2} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.8

Combine $4$ and $\frac{e^{\frac{t}{2}}}{2}$ .

$\frac{4 e^{\frac{t}{2}}}{2} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.9

Cancel the common factor of $4$ and $2$ .

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Step 1.2.9.1

Factor $2$ out of $4 e^{\frac{t}{2}}$ .

$\frac{2 (2 e^{\frac{t}{2}})}{2} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.9.2

Cancel the common factors.

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Step 1.2.9.2.1

Factor $2$ out of $2$ .

$\frac{2 (2 e^{\frac{t}{2}})}{2 (1)} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.9.2.2

Cancel the common factor.

$\frac{2 (2 e^{\frac{t}{2}})}{2 \cdot 1} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.9.2.3

Rewrite the expression.

$\frac{2 e^{\frac{t}{2}}}{1} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.2.9.2.4

Divide $2 e^{\frac{t}{2}}$ by $1$ .

$2 e^{\frac{t}{2}} + \frac{d}{d t} [- t] + \frac{d}{d t} [1]$

Step 1.3

Evaluate $\frac{d}{d t} [- t]$ .

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Step 1.3.1

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

$2 e^{\frac{t}{2}} - \frac{d}{d t} [t] + \frac{d}{d t} [1]$

Step 1.3.2

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

$2 e^{\frac{t}{2}} - 1 \cdot 1 + \frac{d}{d t} [1]$

Step 1.3.3

Multiply $- 1$ by $1$ .

$2 e^{\frac{t}{2}} - 1 + \frac{d}{d t} [1]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$2 e^{\frac{t}{2}} - 1 + 0$

Step 1.4.2

Add $2 e^{\frac{t}{2}} - 1$ and $0$ .

$f' (t) = 2 e^{\frac{t}{2}} - 1$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $2 e^{\frac{t}{2}} - 1$ with respect to $t$ is $\frac{d}{d t} [2 e^{\frac{t}{2}}] + \frac{d}{d t} [- 1]$ .

$\frac{d}{d t} [2 e^{\frac{t}{2}}] + \frac{d}{d t} [- 1]$

Step 2.2

Evaluate $\frac{d}{d t} [2 e^{\frac{t}{2}}]$ .

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Step 2.2.1

Since $2$ is constant with respect to $t$ , the derivative of $2 e^{\frac{t}{2}}$ with respect to $t$ is $2 \frac{d}{d t} [e^{\frac{t}{2}}]$ .

$2 \frac{d}{d t} [e^{\frac{t}{2}}] + \frac{d}{d t} [- 1]$

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = \frac{t}{2}$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $\frac{t}{2}$ .

$2 (\frac{d}{d u} [e^{u}] \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- 1]$

Step 2.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$2 (e^{u} \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- 1]$

Step 2.2.2.3

Replace all occurrences of $u$ with $\frac{t}{2}$ .

$2 (e^{\frac{t}{2}} \frac{d}{d t} [\frac{t}{2}]) + \frac{d}{d t} [- 1]$

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{t}{2}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [t]$ .

$2 (e^{\frac{t}{2}} (\frac{1}{2} \frac{d}{d t} [t])) + \frac{d}{d t} [- 1]$

Step 2.2.4

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

$2 (e^{\frac{t}{2}} (\frac{1}{2} \cdot 1)) + \frac{d}{d t} [- 1]$

Step 2.2.5

Multiply $\frac{1}{2}$ by $1$ .

$2 (e^{\frac{t}{2}} \frac{1}{2}) + \frac{d}{d t} [- 1]$

Step 2.2.6

Combine $e^{\frac{t}{2}}$ and $\frac{1}{2}$ .

$2 \frac{e^{\frac{t}{2}}}{2} + \frac{d}{d t} [- 1]$

Step 2.2.7

Combine $2$ and $\frac{e^{\frac{t}{2}}}{2}$ .

$\frac{2 e^{\frac{t}{2}}}{2} + \frac{d}{d t} [- 1]$

Step 2.2.8

Cancel the common factor of $2$ .

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Step 2.2.8.1

Cancel the common factor.

$\frac{2 e^{\frac{t}{2}}}{2} + \frac{d}{d t} [- 1]$

Step 2.2.8.2

Divide $e^{\frac{t}{2}}$ by $1$ .

$e^{\frac{t}{2}} + \frac{d}{d t} [- 1]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- 1$ with respect to $t$ is $0$ .

$e^{\frac{t}{2}} + 0$

Step 2.3.2

Add $e^{\frac{t}{2}}$ and $0$ .

$f'' (t) = e^{\frac{t}{2}}$

Step 3

Find the third derivative.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = \frac{t}{2}$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $\frac{t}{2}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d t} [\frac{t}{2}]$

Step 3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d t} [\frac{t}{2}]$

Step 3.1.3

Replace all occurrences of $u$ with $\frac{t}{2}$ .

$e^{\frac{t}{2}} \frac{d}{d t} [\frac{t}{2}]$

Step 3.2

Differentiate.

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Step 3.2.1

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{t}{2}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [t]$ .

$e^{\frac{t}{2}} (\frac{1}{2} \frac{d}{d t} [t])$

Step 3.2.2

Combine $\frac{1}{2}$ and $e^{\frac{t}{2}}$ .

$\frac{e^{\frac{t}{2}}}{2} \frac{d}{d t} [t]$

Step 3.2.3

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

$\frac{e^{\frac{t}{2}}}{2} \cdot 1$

Step 3.2.4

Multiply $\frac{e^{\frac{t}{2}}}{2}$ by $1$ .

$f''' (t) = \frac{e^{\frac{t}{2}}}{2}$

Step 4

Find the fourth derivative.

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Step 4.1

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{e^{\frac{t}{2}}}{2}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [e^{\frac{t}{2}}]$ .

$\frac{1}{2} \frac{d}{d t} [e^{\frac{t}{2}}]$

Step 4.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = \frac{t}{2}$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $\frac{t}{2}$ .

$\frac{1}{2} (\frac{d}{d u} [e^{u}] \frac{d}{d t} [\frac{t}{2}])$

Step 4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{1}{2} (e^{u} \frac{d}{d t} [\frac{t}{2}])$

Step 4.2.3

Replace all occurrences of $u$ with $\frac{t}{2}$ .

$\frac{1}{2} (e^{\frac{t}{2}} \frac{d}{d t} [\frac{t}{2}])$

Step 4.3

Differentiate.

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Step 4.3.1

Combine $e^{\frac{t}{2}}$ and $\frac{1}{2}$ .

$\frac{e^{\frac{t}{2}}}{2} \frac{d}{d t} [\frac{t}{2}]$

Step 4.3.2

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{t}{2}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [t]$ .

$\frac{e^{\frac{t}{2}}}{2} (\frac{1}{2} \frac{d}{d t} [t])$

Step 4.3.3

Combine fractions.

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Step 4.3.3.1

Multiply $\frac{1}{2}$ by $\frac{e^{\frac{t}{2}}}{2}$ .

$\frac{e^{\frac{t}{2}}}{2 \cdot 2} \frac{d}{d t} [t]$

Step 4.3.3.2

Multiply $2$ by $2$ .

$\frac{e^{\frac{t}{2}}}{4} \frac{d}{d t} [t]$

Step 4.3.4

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

$\frac{e^{\frac{t}{2}}}{4} \cdot 1$

Step 4.3.5

Multiply $\frac{e^{\frac{t}{2}}}{4}$ by $1$ .

$f^{4} (t) = \frac{e^{\frac{t}{2}}}{4}$

Step 5

The fourth derivative of $P (t)$ with respect to $t$ is $\frac{e^{\frac{t}{2}}}{4}$ .

$\frac{e^{\frac{t}{2}}}{4}$