What is the d/(dt)(2t^{-1}) ?

The d/(dt)(2t^{-1}) is -2/(t^2)

What is the first d/(dt)(2t^{-1}) ?

The first d/(dt)(2t^{-1}) is -2/(t^2)

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\frac{d}{d t} (2 t^{- 1})

- \frac{2}{t ^{2}}

Solution steps

\frac{d}{d t} (2 t^{- 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d t} (t^{- 1})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 (- 1 \cdot t^{- 1 - 1})

Simplify

2 (- 1 \cdot t^{- 1 - 1}) : - \frac{2}{t ^{2}}

= - \frac{2}{t ^{2}}