What is the d/(dt)(5t^{-3}) ?

The d/(dt)(5t^{-3}) is -(15)/(t^4)

What is the first d/(dt)(5t^{-3}) ?

The first d/(dt)(5t^{-3}) is -(15)/(t^4)

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\frac{d}{d t} (5 t^{- 3})

- \frac{15}{t ^{4}}

Solution steps

\frac{d}{d t} (5 t^{- 3})

Take the constant out:

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

= 5 \frac{d}{d t} (t^{- 3})

Apply the Power Rule:

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

= 5 (- 3 t^{- 3 - 1})

Simplify

5 (- 3 t^{- 3 - 1}) : - \frac{15}{t ^{4}}

= - \frac{15}{t ^{4}}

\int_{- 1}^{1} e^{y} - y^{2} + 2 d y

\frac{\partial}{\partial y} (tan (3 x - 2 y + 1))

\frac{s + 7}{s ^{2} + 8 s + 25}

g (x) = (3 - 4 x)^{3}

\frac{d}{d x} (\frac{10 x}{x ^{2} + 4})