Calculus Examples

$g (t) = {(\frac{t - 2}{2 t + 1})}^{9}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{9}$ and $g (t) = \frac{t - 2}{2 t + 1}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{t - 2}{2 t + 1}$ .

$\frac{d}{d u} [u^{9}] \frac{d}{d t} [\frac{t - 2}{2 t + 1}]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 9$ .

$9 u^{8} \frac{d}{d t} [\frac{t - 2}{2 t + 1}]$

Step 1.3

Replace all occurrences of $u$ with $\frac{t - 2}{2 t + 1}$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{d}{d t} [\frac{t - 2}{2 t + 1}]$

Step 2

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 $f (t) = t - 2$ and $g (t) = 2 t + 1$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{(2 t + 1) \frac{d}{d t} [t - 2] - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3

Differentiate.

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Step 3.1

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{(2 t + 1) (\frac{d}{d t} [t] + \frac{d}{d t} [- 2]) - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3.2

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{(2 t + 1) (1 + \frac{d}{d t} [- 2]) - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3.3

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{(2 t + 1) (1 + 0) - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{(2 t + 1) \cdot 1 - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3.4.2

Multiply $2 t + 1$ by $1$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) \frac{d}{d t} [2 t + 1]}{{(2 t + 1)}^{2}}$

Step 3.5

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) (\frac{d}{d t} [2 t] + \frac{d}{d t} [1])}{{(2 t + 1)}^{2}}$

Step 3.6

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) (2 \frac{d}{d t} [t] + \frac{d}{d t} [1])}{{(2 t + 1)}^{2}}$

Step 3.7

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) (2 \cdot 1 + \frac{d}{d t} [1])}{{(2 t + 1)}^{2}}$

Step 3.8

Multiply $2$ by $1$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) (2 + \frac{d}{d t} [1])}{{(2 t + 1)}^{2}}$

Step 3.9

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

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) (2 + 0)}{{(2 t + 1)}^{2}}$

Step 3.10

Combine fractions.

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Step 3.10.1

Add $2$ and $0$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - (t - 2) \cdot 2}{{(2 t + 1)}^{2}}$

Step 3.10.2

Multiply $2$ by $- 1$ .

$9 {(\frac{t - 2}{2 t + 1})}^{8} \frac{2 t + 1 - 2 (t - 2)}{{(2 t + 1)}^{2}}$

Step 3.10.3

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

$\frac{(2 t + 1 - 2 (t - 2)) \cdot 9}{{(2 t + 1)}^{2}} {(\frac{t - 2}{2 t + 1})}^{8}$

Step 3.10.4

Move $9$ to the left of $2 t + 1 - 2 (t - 2)$ .

$\frac{9 \cdot (2 t + 1 - 2 (t - 2))}{{(2 t + 1)}^{2}} {(\frac{t - 2}{2 t + 1})}^{8}$

Step 4

Simplify.

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Step 4.1

Apply the product rule to $\frac{t - 2}{2 t + 1}$ .

$\frac{9 (2 t + 1 - 2 (t - 2))}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.2

Apply the distributive property.

$\frac{9 (2 t + 1 - 2 t - 2 \cdot - 2)}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.3

Apply the distributive property.

$\frac{9 (2 t) + 9 \cdot 1 + 9 (- 2 t) + 9 (- 2 \cdot - 2)}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4

Combine terms.

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Step 4.4.1

Multiply $2$ by $9$ .

$\frac{18 t + 9 \cdot 1 + 9 (- 2 t) + 9 (- 2 \cdot - 2)}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.2

Multiply $9$ by $1$ .

$\frac{18 t + 9 + 9 (- 2 t) + 9 (- 2 \cdot - 2)}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.3

Multiply $- 2$ by $9$ .

$\frac{18 t + 9 - 18 t + 9 (- 2 \cdot - 2)}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.4

Multiply $- 2$ by $- 2$ .

$\frac{18 t + 9 - 18 t + 9 \cdot 4}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.5

Multiply $9$ by $4$ .

$\frac{18 t + 9 - 18 t + 36}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.6

Subtract $18 t$ from $18 t$ .

$\frac{0 + 9 + 36}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.7

Add $0$ and $9$ .

$\frac{9 + 36}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.8

Add $9$ and $36$ .

$\frac{45}{{(2 t + 1)}^{2}} \cdot \frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$

Step 4.4.9

Multiply $\frac{45}{{(2 t + 1)}^{2}}$ by $\frac{{(t - 2)}^{8}}{{(2 t + 1)}^{8}}$ .

$\frac{45 {(t - 2)}^{8}}{{(2 t + 1)}^{2} {(2 t + 1)}^{8}}$

Step 4.4.10

Multiply ${(2 t + 1)}^{2}$ by ${(2 t + 1)}^{8}$ by adding the exponents.

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Step 4.4.10.1

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

$\frac{45 {(t - 2)}^{8}}{{(2 t + 1)}^{2 + 8}}$

Step 4.4.10.2

Add $2$ and $8$ .

$\frac{45 {(t - 2)}^{8}}{{(2 t + 1)}^{10}}$