Calculus Examples

$\frac{\cos (6 t)}{t^{3} - 27}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = \cos (6 t)$ and $g (t) = t^{3} - 27$ .

$\frac{(t^{3} - 27) \frac{d}{d t} [\cos (6 t)] - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = 6 t$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $6 t$ .

$\frac{(t^{3} - 27) (\frac{d}{d u} [\cos (u)] \frac{d}{d t} [6 t]) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 2.2

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

$\frac{(t^{3} - 27) (- \sin (u) \frac{d}{d t} [6 t]) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 2.3

Replace all occurrences of $u$ with $6 t$ .

$\frac{(t^{3} - 27) (- \sin (6 t) \frac{d}{d t} [6 t]) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 3

Differentiate.

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Step 3.1

Since $6$ is constant with respect to $t$ , the derivative of $6 t$ with respect to $t$ is $6 \frac{d}{d t} [t]$ .

$\frac{(t^{3} - 27) (- \sin (6 t) (6 \frac{d}{d t} [t])) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 3.2

Multiply $6$ by $- 1$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t) \frac{d}{d t} [t]) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 3.3

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

$\frac{(t^{3} - 27) (- 6 \sin (6 t) \cdot 1) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 3.4

Multiply $- 6$ by $1$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - \cos (6 t) \frac{d}{d t} [t^{3} - 27]}{{(t^{3} - 27)}^{2}}$

Step 3.5

By the Sum Rule, the derivative of $t^{3} - 27$ with respect to $t$ is $\frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 27]$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - \cos (6 t) (\frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 27])}{{(t^{3} - 27)}^{2}}$

Step 3.6

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

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - \cos (6 t) (3 t^{2} + \frac{d}{d t} [- 27])}{{(t^{3} - 27)}^{2}}$

Step 3.7

Since $- 27$ is constant with respect to $t$ , the derivative of $- 27$ with respect to $t$ is $0$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - \cos (6 t) (3 t^{2} + 0)}{{(t^{3} - 27)}^{2}}$

Step 3.8

Simplify the expression.

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Step 3.8.1

Add $3 t^{2}$ and $0$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - \cos (6 t) (3 t^{2})}{{(t^{3} - 27)}^{2}}$

Step 3.8.2

Multiply $3$ by $- 1$ .

$\frac{(t^{3} - 27) (- 6 \sin (6 t)) - 3 \cos (6 t) t^{2}}{{(t^{3} - 27)}^{2}}$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$\frac{t^{3} (- 6 \sin (6 t)) - 27 (- 6 \sin (6 t)) - 3 \cos (6 t) t^{2}}{{(t^{3} - 27)}^{2}}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{- 6 t^{3} \sin (6 t) - 27 (- 6 \sin (6 t)) - 3 \cos (6 t) t^{2}}{{(t^{3} - 27)}^{2}}$

Step 4.2.1.2

Multiply $- 6$ by $- 27$ .

$\frac{- 6 t^{3} \sin (6 t) + 162 \sin (6 t) - 3 \cos (6 t) t^{2}}{{(t^{3} - 27)}^{2}}$

Step 4.2.2

Reorder factors in $- 6 t^{3} \sin (6 t) + 162 \sin (6 t) - 3 \cos (6 t) t^{2}$ .

$\frac{- 6 t^{3} \sin (6 t) + 162 \sin (6 t) - 3 t^{2} \cos (6 t)}{{(t^{3} - 27)}^{2}}$

Step 4.3

Reorder terms.

$\frac{- 6 t^{3} \sin (6 t) - 3 t^{2} \cos (6 t) + 162 \sin (6 t)}{{(t^{3} - 27)}^{2}}$