Calculus Examples

$h (t) = {(6 t^{2} - 5 t)}^{- 3}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 3}$ and $g (t) = 6 t^{2} - 5 t$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $6 t^{2} - 5 t$ .

$\frac{d}{d u} [u^{- 3}] \frac{d}{d t} [6 t^{2} - 5 t]$

Step 1.2

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

$- 3 u^{- 4} \frac{d}{d t} [6 t^{2} - 5 t]$

Step 1.3

Replace all occurrences of $u$ with $6 t^{2} - 5 t$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} \frac{d}{d t} [6 t^{2} - 5 t]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $6 t^{2} - 5 t$ with respect to $t$ is $\frac{d}{d t} [6 t^{2}] + \frac{d}{d t} [- 5 t]$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (\frac{d}{d t} [6 t^{2}] + \frac{d}{d t} [- 5 t])$

Step 2.2

Since $6$ is constant with respect to $t$ , the derivative of $6 t^{2}$ with respect to $t$ is $6 \frac{d}{d t} [t^{2}]$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (6 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 5 t])$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (6 (2 t) + \frac{d}{d t} [- 5 t])$

Step 2.4

Multiply $2$ by $6$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (12 t + \frac{d}{d t} [- 5 t])$

Step 2.5

Since $- 5$ is constant with respect to $t$ , the derivative of $- 5 t$ with respect to $t$ is $- 5 \frac{d}{d t} [t]$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (12 t - 5 \frac{d}{d t} [t])$

Step 2.6

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

$- 3 {(6 t^{2} - 5 t)}^{- 4} (12 t - 5 \cdot 1)$

Step 2.7

Multiply $- 5$ by $1$ .

$- 3 {(6 t^{2} - 5 t)}^{- 4} (12 t - 5)$

Step 3

Simplify.

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Step 3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 3 \frac{1}{{(6 t^{2} - 5 t)}^{4}} (12 t - 5)$

Step 3.2

Combine terms.

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Step 3.2.1

Combine $- 3$ and $\frac{1}{{(6 t^{2} - 5 t)}^{4}}$ .

$\frac{- 3}{{(6 t^{2} - 5 t)}^{4}} (12 t - 5)$

Step 3.2.2

Move the negative in front of the fraction.

$- \frac{3}{{(6 t^{2} - 5 t)}^{4}} (12 t - 5)$

Step 3.3

Reorder the factors of $- \frac{3}{{(6 t^{2} - 5 t)}^{4}} (12 t - 5)$ .

$- (12 t - 5) \frac{3}{{(6 t^{2} - 5 t)}^{4}}$

Step 3.4

Apply the distributive property.

$(- (12 t) - - 5) \frac{3}{{(6 t^{2} - 5 t)}^{4}}$

Step 3.5

Multiply $12$ by $- 1$ .

$(- 12 t - - 5) \frac{3}{{(6 t^{2} - 5 t)}^{4}}$

Step 3.6

Multiply $- 1$ by $- 5$ .

$(- 12 t + 5) \frac{3}{{(6 t^{2} - 5 t)}^{4}}$

Step 3.7

Simplify the denominator.

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Step 3.7.1

Factor $t$ out of $6 t^{2} - 5 t$ .

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Step 3.7.1.1

Factor $t$ out of $6 t^{2}$ .

$(- 12 t + 5) \frac{3}{{(t (6 t) - 5 t)}^{4}}$

Step 3.7.1.2

Factor $t$ out of $- 5 t$ .

$(- 12 t + 5) \frac{3}{{(t (6 t) + t \cdot - 5)}^{4}}$

Step 3.7.1.3

Factor $t$ out of $t (6 t) + t \cdot - 5$ .

$(- 12 t + 5) \frac{3}{{(t (6 t - 5))}^{4}}$

Step 3.7.2

Apply the product rule to $t (6 t - 5)$ .

$(- 12 t + 5) \frac{3}{t^{4} {(6 t - 5)}^{4}}$

Step 3.8

Multiply $- 12 t + 5$ by $\frac{3}{t^{4} {(6 t - 5)}^{4}}$ .

$\frac{(- 12 t + 5) \cdot 3}{t^{4} {(6 t - 5)}^{4}}$

Step 3.9

Move $3$ to the left of $- 12 t + 5$ .

$\frac{3 \cdot (- 12 t + 5)}{t^{4} {(6 t - 5)}^{4}}$

Step 3.10

Factor $- 1$ out of $- 12 t$ .

$\frac{3 (- (12 t) + 5)}{t^{4} {(6 t - 5)}^{4}}$

Step 3.11

Rewrite $5$ as $- 1 (- 5)$ .

$\frac{3 (- (12 t) - 1 (- 5))}{t^{4} {(6 t - 5)}^{4}}$

Step 3.12

Factor $- 1$ out of $- (12 t) - 1 (- 5)$ .

$\frac{3 (- (12 t - 5))}{t^{4} {(6 t - 5)}^{4}}$

Step 3.13

Rewrite $- (12 t - 5)$ as $- 1 (12 t - 5)$ .

$\frac{3 (- 1 (12 t - 5))}{t^{4} {(6 t - 5)}^{4}}$

Step 3.14

Move the negative in front of the fraction.

$- \frac{3 (12 t - 5)}{t^{4} {(6 t - 5)}^{4}}$