Calculus Examples

$\frac{1}{1 + t}$

Step 1

Find the first derivative.

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Step 1.1

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

$f' (t) = \frac{d}{d t} ({(1 + t)}^{- 1})$

Step 1.2

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 $g (t) = 1 + t$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $1 + t$ .

$f' (t) = \frac{d}{d u} (u^{- 1}) \frac{d}{d t} (1 + t)$

Step 1.2.2

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

$f' (t) = - u^{- 2} \frac{d}{d t} (1 + t)$

Step 1.2.3

Replace all occurrences of $u$ with $1 + t$ .

$f' (t) = - {(1 + t)}^{- 2} \frac{d}{d t} (1 + t)$

Step 1.3

Differentiate.

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Step 1.3.1

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

$f' (t) = - {(1 + t)}^{- 2} (\frac{d}{d t} (1) + \frac{d}{d t} (t))$

Step 1.3.2

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

$f' (t) = - {(1 + t)}^{- 2} (0 + \frac{d}{d t} (t))$

Step 1.3.3

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

$f' (t) = - {(1 + t)}^{- 2} \frac{d}{d t} t$

Step 1.3.4

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

$f' (t) = - {(1 + t)}^{- 2} \cdot 1$

Step 1.3.5

Multiply $- 1$ by $1$ .

$f' (t) = - {(1 + t)}^{- 2}$

Step 1.4

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

$f' (t) = - \frac{1}{{(1 + t)}^{2}}$

Step 2

Find the second derivative.

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Step 2.1

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

$f'' (t) = - \frac{d}{d t} \cdot \frac{1}{{(1 + t)}^{2}} + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.2

Apply basic rules of exponents.

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Step 2.2.1

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

$f'' (t) = - \frac{d}{d t} {({(1 + t)}^{2})}^{- 1} + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (t) = - \frac{d}{d t} {(1 + t)}^{2 \cdot - 1} + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.2.2.2

Multiply $2$ by $- 1$ .

$f'' (t) = - \frac{d}{d t} {(1 + t)}^{- 2} + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.3

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 $g (t) = 1 + t$ .

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Step 2.3.1

To apply the Chain Rule, set $u$ as $1 + t$ .

$f'' (t) = - (\frac{d}{d u} (u^{- 2}) \frac{d}{d t} (1 + t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.3.2

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

$f'' (t) = - (- 2 u^{- 3} \frac{d}{d t} (1 + t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.3.3

Replace all occurrences of $u$ with $1 + t$ .

$f'' (t) = - (- 2 {(1 + t)}^{- 3} \frac{d}{d t} (1 + t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4

Differentiate.

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Step 2.4.1

Multiply $- 2$ by $- 1$ .

$f'' (t) = 2 ({(1 + t)}^{- 3} \frac{d}{d t} (1 + t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.2

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

$f'' (t) = 2 {(1 + t)}^{- 3} (\frac{d}{d t} (1) + \frac{d}{d t} (t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.3

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

$f'' (t) = 2 {(1 + t)}^{- 3} (0 + \frac{d}{d t} (t)) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.4

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

$f'' (t) = 2 {(1 + t)}^{- 3} \frac{d}{d t} (t) + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.5

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

$f'' (t) = 2 {(1 + t)}^{- 3} \cdot 1 + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.6

Multiply $2$ by $1$ .

$f'' (t) = 2 {(1 + t)}^{- 3} + \frac{1}{{(1 + t)}^{2}} \cdot \frac{d}{d t} (- 1)$

Step 2.4.7

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

$f'' (t) = 2 {(1 + t)}^{- 3} + \frac{1}{{(1 + t)}^{2}} \cdot 0$

Step 2.4.8

Simplify the expression.

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Step 2.4.8.1

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

$f'' (t) = 2 {(1 + t)}^{- 3} + 0$

Step 2.4.8.2

Add $2 {(1 + t)}^{- 3}$ and $0$ .

$f'' (t) = 2 {(1 + t)}^{- 3}$

Step 2.5

Simplify.

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Step 2.5.1

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

$f'' (t) = 2 (\frac{1}{{(1 + t)}^{3}})$

Step 2.5.2

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

$f'' (t) = \frac{2}{{(1 + t)}^{3}}$

Step 2.5.3

Reorder terms.

$f'' (t) = \frac{2}{{(t + 1)}^{3}}$