What is the derivative of g(t)= 9/(sqrt(t)) ?

The derivative of g(t)= 9/(sqrt(t)) is -9/(2t^{3/2)}

What is the first derivative of g(t)= 9/(sqrt(t)) ?

The first derivative of g(t)= 9/(sqrt(t)) is -9/(2t^{3/2)}

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derivative of

g (t) = \frac{9}{t}

- \frac{9}{2 t ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{9}{t})

Take the constant out:

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

= 9 \frac{d}{d t} (\frac{1}{t})

Apply exponent rule:

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

= 9 \frac{d}{d t} (t^{- \frac{1}{2}})

Apply the Power Rule:

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

= 9 (- \frac{1}{2} t^{- \frac{1}{2} - 1})

Simplify

9 (- \frac{1}{2} t^{- \frac{1}{2} - 1}) : - \frac{9}{2 t ^{\frac{3}{2}}}

= - \frac{9}{2 t ^{\frac{3}{2}}}

What is the derivative of g(t)= 9/(sqrt(t)) ?
The derivative of g(t)= 9/(sqrt(t)) is -9/(2t^{3/2)}
What is the first derivative of g(t)= 9/(sqrt(t)) ?
The first derivative of g(t)= 9/(sqrt(t)) is -9/(2t^{3/2)}