What is the derivative of g(t)=4t^{-3/8} ?

The derivative of g(t)=4t^{-3/8} is -3/(2t^{11/8)}

What is the first derivative of g(t)=4t^{-3/8} ?

The first derivative of g(t)=4t^{-3/8} is -3/(2t^{11/8)}

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derivative of

g (t) = 4 t^{- \frac{3}{8}}

- \frac{3}{2 t ^{\frac{11}{8}}}

Solution steps

\frac{d}{d t} (4 t^{- \frac{3}{8}})

Take the constant out:

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

= 4 \frac{d}{d t} (t^{- \frac{3}{8}})

Apply the Power Rule:

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

= 4 (- \frac{3}{8} t^{- \frac{3}{8} - 1})

Simplify

4 (- \frac{3}{8} t^{- \frac{3}{8} - 1}) : - \frac{3}{2 t ^{\frac{11}{8}}}

= - \frac{3}{2 t ^{\frac{11}{8}}}

What is the derivative of g(t)=4t^{-3/8} ?
The derivative of g(t)=4t^{-3/8} is -3/(2t^{11/8)}
What is the first derivative of g(t)=4t^{-3/8} ?
The first derivative of g(t)=4t^{-3/8} is -3/(2t^{11/8)}