Calculus Examples

$f (t) = \frac{1}{\arctan (3 t)}$

Step 1

Rewrite $\frac{1}{\arctan (3 t)}$ as ${\arctan (3 t)}^{- 1}$ .

$\frac{d}{d t} [{\arctan (3 t)}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = \arctan (3 t)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $\arctan (3 t)$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $\arctan (3 t)$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

The derivative of $\arctan (u_{2})$ with respect to $u_{2}$ is $\frac{1}{1 + {u_{2}}^{2}}$ .

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

Step 3.3

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

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

Step 4

Differentiate.

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Step 4.1

Factor $3$ out of $3 t$ .

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

Step 4.2

Combine fractions.

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Step 4.2.1

Simplify the expression.

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Step 4.2.1.1

Apply the product rule to $3 (t)$ .

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

Step 4.2.1.2

Raise $3$ to the power of $2$ .

$- {\arctan (3 t)}^{- 2} (\frac{1}{1 + 9 t^{2}} \frac{d}{d t} [3 t])$

Step 4.2.2

Combine $\frac{1}{1 + 9 t^{2}}$ and ${\arctan (3 t)}^{- 2}$ .

$- \frac{{\arctan (3 t)}^{- 2}}{1 + 9 t^{2}} \frac{d}{d t} [3 t]$

Step 4.2.3

Move ${\arctan (3 t)}^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} \frac{d}{d t} [3 t]$

Step 4.3

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

$- \frac{1}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} (3 \frac{d}{d t} [t])$

Step 4.4

Combine fractions.

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Step 4.4.1

Multiply $3$ by $- 1$ .

$- 3 \frac{1}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} \frac{d}{d t} [t]$

Step 4.4.2

Combine $- 3$ and $\frac{1}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}}$ .

$\frac{- 3}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} \frac{d}{d t} [t]$

Step 4.4.3

Move the negative in front of the fraction.

$- \frac{3}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} \frac{d}{d t} [t]$

Step 4.5

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

$- \frac{3}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}} \cdot 1$

Step 4.6

Multiply $- 1$ by $1$ .

$- \frac{3}{(1 + 9 t^{2}) {\arctan (3 t)}^{2}}$

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

$- \frac{3}{1 {\arctan (3 t)}^{2} + 9 t^{2} {\arctan (3 t)}^{2}}$

Step 5.2

Multiply ${\arctan (3 t)}^{2}$ by $1$ .

$- \frac{3}{{\arctan (3 t)}^{2} + 9 t^{2} {\arctan (3 t)}^{2}}$

Step 5.3

Factor ${\arctan (3 t)}^{2}$ out of ${\arctan (3 t)}^{2} + 9 t^{2} {\arctan (3 t)}^{2}$ .

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Step 5.3.1

Multiply by $1$ .

$- \frac{3}{{\arctan (3 t)}^{2} \cdot 1 + 9 t^{2} {\arctan (3 t)}^{2}}$

Step 5.3.2

Factor ${\arctan (3 t)}^{2}$ out of $9 t^{2} {\arctan (3 t)}^{2}$ .

$- \frac{3}{{\arctan (3 t)}^{2} \cdot 1 + {\arctan (3 t)}^{2} (9 t^{2})}$

Step 5.3.3

Factor ${\arctan (3 t)}^{2}$ out of ${\arctan (3 t)}^{2} \cdot 1 + {\arctan (3 t)}^{2} (9 t^{2})$ .

$- \frac{3}{{\arctan (3 t)}^{2} (1 + 9 t^{2})}$