Calculus Examples

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

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = - 1$ and $g (t) = \frac{1}{{(3 + t)}^{2}}$ .

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

Step 2

Apply basic rules of exponents.

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Step 2.1

Rewrite $\frac{1}{{(3 + t)}^{2}}$ as ${({(3 + t)}^{2})}^{- 1}$ .

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

Step 2.2

Multiply the exponents in ${({(3 + t)}^{2})}^{- 1}$ .

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Step 2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.2.2

Multiply $2$ by $- 1$ .

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

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u$ as $3 + t$ .

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

Step 3.2

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

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

Step 3.3

Replace all occurrences of $u$ with $3 + t$ .

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

Step 4

Differentiate.

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Step 4.1

Multiply $- 2$ by $- 1$ .

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

Step 4.2

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

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

Step 4.3

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$2 {(3 + t)}^{- 3} (0 + \frac{d}{d t} [t]) + \frac{1}{{(3 + t)}^{2}} \frac{d}{d t} [- 1]$

Step 4.4

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

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

Step 4.5

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

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

Step 4.6

Multiply $2$ by $1$ .

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

Step 4.7

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

$2 {(3 + t)}^{- 3} + \frac{1}{{(3 + t)}^{2}} \cdot 0$

Step 4.8

Simplify the expression.

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Step 4.8.1

Multiply $\frac{1}{{(3 + t)}^{2}}$ by $0$ .

$2 {(3 + t)}^{- 3} + 0$

Step 4.8.2

Add $2 {(3 + t)}^{- 3}$ and $0$ .

$2 {(3 + t)}^{- 3}$

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Reorder terms.

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