What is the d/(dt)(2t^{-3/4}) ?

The d/(dt)(2t^{-3/4}) is -3/(2t^{7/4)}

What is the first d/(dt)(2t^{-3/4}) ?

The first d/(dt)(2t^{-3/4}) is -3/(2t^{7/4)}

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\frac{d}{d t} (2 t^{- \frac{3}{4}})

- \frac{3}{2 t ^{\frac{7}{4}}}

Solution steps

\frac{d}{d t} (2 t^{- \frac{3}{4}})

Take the constant out:

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

= 2 \frac{d}{d t} (t^{- \frac{3}{4}})

Apply the Power Rule:

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

= 2 (- \frac{3}{4} t^{- \frac{3}{4} - 1})

Simplify

2 (- \frac{3}{4} t^{- \frac{3}{4} - 1}) : - \frac{3}{2 t ^{\frac{7}{4}}}

= - \frac{3}{2 t ^{\frac{7}{4}}}

x \to \infty lim ((e)^{x})

\frac{\partial}{\partial x} (x^{2} (\frac{y}{x}))

(f (x) (x)^{g (x)})^{'}

x \to 2 lim (\frac{4 x ^{2} - 2 x + 5}{x})

\int \frac{435}{3 x + 14} - 15 d x

What is the d/(dt)(2t^{-3/4}) ?
The d/(dt)(2t^{-3/4}) is -3/(2t^{7/4)}
What is the first d/(dt)(2t^{-3/4}) ?
The first d/(dt)(2t^{-3/4}) is -3/(2t^{7/4)}