What is the derivative of 9/(t^3) ?

The derivative of 9/(t^3) is -(27)/(t^4)

What is the first derivative of 9/(t^3) ?

The first derivative of 9/(t^3) is -(27)/(t^4)

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derivative of

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

- \frac{27}{t ^{4}}

Solution steps

\frac{d}{d t} (\frac{9}{t ^{3}})

Take the constant out:

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

= 9 \frac{d}{d t} (\frac{1}{t ^{3}})

Apply exponent rule:

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

= 9 \frac{d}{d t} (t^{- 3})

Apply the Power Rule:

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

= 9 (- 3 t^{- 3 - 1})

Simplify

9 (- 3 t^{- 3 - 1}) : - \frac{27}{t ^{4}}

= - \frac{27}{t ^{4}}

\int_{- π}^{π} t sin (n t) d t

k = 0 \sum \infty x^{k + 1}

im pl i c i t \frac{d y}{d x}, x y = 9

in v erse l a pl a ce \frac{1}{s - 8}

\int_{0}^{1} \frac{2}{2 - x ^{2}} d x

What is the derivative of 9/(t^3) ?
The derivative of 9/(t^3) is -(27)/(t^4)
What is the first derivative of 9/(t^3) ?
The first derivative of 9/(t^3) is -(27)/(t^4)