Calculus Examples

$- \frac{1}{t}$

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

$- \frac{d}{d t} [\frac{1}{t}] + \frac{1}{t} \frac{d}{d t} [- 1]$

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Multiply.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.2

Multiply $t^{- 2}$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify the expression.

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Step 2.5.1

Multiply $\frac{1}{t}$ by $0$ .

$t^{- 2} + 0$

Step 2.5.2

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

$t^{- 2}$

Step 3

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

$\frac{1}{t^{2}}$