Calculus Examples

$y = \ln (| 2 + t - t^{3} |)$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = | 2 + t - t^{3} |$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $| 2 + t - t^{3} |$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d t} [| 2 + t - t^{3} |]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d t} [| 2 + t - t^{3} |]$

Step 1.3

Replace all occurrences of $u_{1}$ with $| 2 + t - t^{3} |$ .

$\frac{1}{| 2 + t - t^{3} |} \frac{d}{d t} [| 2 + t - t^{3} |]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = | t |$ and $g (t) = 2 + t - t^{3}$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $2 + t - t^{3}$ .

$\frac{1}{| 2 + t - t^{3} |} (\frac{d}{d u_{2}} [| u_{2} |] \frac{d}{d t} [2 + t - t^{3}])$

Step 2.2

The derivative of $| u_{2} |$ with respect to $u_{2}$ is $\frac{u_{2}}{| u_{2} |}$ .

$\frac{1}{| 2 + t - t^{3} |} (\frac{u_{2}}{| u_{2} |} \frac{d}{d t} [2 + t - t^{3}])$

Step 2.3

Replace all occurrences of $u_{2}$ with $2 + t - t^{3}$ .

$\frac{1}{| 2 + t - t^{3} |} (\frac{2 + t - t^{3}}{| 2 + t - t^{3} |} \frac{d}{d t} [2 + t - t^{3}])$

Step 3

Multiply $\frac{2 + t - t^{3}}{| 2 + t - t^{3} |}$ by $\frac{1}{| 2 + t - t^{3} |}$ .

$\frac{2 + t - t^{3}}{| 2 + t - t^{3} | | 2 + t - t^{3} |} \frac{d}{d t} [2 + t - t^{3}]$

Step 4

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{2 + t - t^{3}}{| (2 + t - t^{3}) (2 + t - t^{3}) |} \frac{d}{d t} [2 + t - t^{3}]$

Step 5

Raise $2 + t - t^{3}$ to the power of $1$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{1} (2 + t - t^{3}) |} \frac{d}{d t} [2 + t - t^{3}]$

Step 6

Raise $2 + t - t^{3}$ to the power of $1$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{1} {(2 + t - t^{3})}^{1} |} \frac{d}{d t} [2 + t - t^{3}]$

Step 7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{1 + 1} |} \frac{d}{d t} [2 + t - t^{3}]$

Step 8

Add $1$ and $1$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} \frac{d}{d t} [2 + t - t^{3}]$

Step 9

By the Sum Rule, the derivative of $2 + t - t^{3}$ with respect to $t$ is $\frac{d}{d t} [2] + \frac{d}{d t} [t] + \frac{d}{d t} [- t^{3}]$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (\frac{d}{d t} [2] + \frac{d}{d t} [t] + \frac{d}{d t} [- t^{3}])$

Step 10

Since $2$ is constant with respect to $t$ , the derivative of $2$ with respect to $t$ is $0$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (0 + \frac{d}{d t} [t] + \frac{d}{d t} [- t^{3}])$

Step 11

Add $0$ and $\frac{d}{d t} [t]$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (\frac{d}{d t} [t] + \frac{d}{d t} [- t^{3}])$

Step 12

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

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (1 + \frac{d}{d t} [- t^{3}])$

Step 13

Since $- 1$ is constant with respect to $t$ , the derivative of $- t^{3}$ with respect to $t$ is $- \frac{d}{d t} [t^{3}]$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (1 - \frac{d}{d t} [t^{3}])$

Step 14

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 3$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (1 - (3 t^{2}))$

Step 15

Simplify the expression.

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Step 15.1

Multiply $3$ by $- 1$ .

$\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (1 - 3 t^{2})$

Step 15.2

Reorder the factors of $\frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |} (1 - 3 t^{2})$ .

$(1 - 3 t^{2}) \frac{2 + t - t^{3}}{| {(2 + t - t^{3})}^{2} |}$