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

The d/(dt)(-1/(t^2)) is 2/(t^3)

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

The first d/(dt)(-1/(t^2)) is 2/(t^3)

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

\frac{2}{t ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

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

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

Simplify

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

= \frac{2}{t ^{3}}

x \to 2 lim (4 - x^{2})

\frac{d}{d x} ((A x^{2} + B x + C) e^{x})

\frac{d}{d x} (3 x - 4)

x \to \infty lim (\frac{e ^{3 x - 8}}{5 x})

\int (- x + 3) d x