What is the derivative of f(t)= 9/(t^2) ?

The derivative of f(t)= 9/(t^2) is -(18)/(t^3)

What is the first derivative of f(t)= 9/(t^2) ?

The first derivative of f(t)= 9/(t^2) is -(18)/(t^3)

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derivative of

f (t) = \frac{9}{t ^{2}}

- \frac{18}{t ^{3}}

Solution steps

\frac{d}{d t} (\frac{9}{t ^{2}})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

9 (- 2 t^{- 2 - 1}) : - \frac{18}{t ^{3}}

= - \frac{18}{t ^{3}}

x (y^{^{'}}) = 4 y

t an g e n t x^{3} + 2

\int 4 ln (x^{2} - 1) d x

\int 2 x^{2} - 4 x - 6 d x

\frac{d}{d x} (x^{\frac{9}{x}})

What is the derivative of f(t)= 9/(t^2) ?
The derivative of f(t)= 9/(t^2) is -(18)/(t^3)
What is the first derivative of f(t)= 9/(t^2) ?
The first derivative of f(t)= 9/(t^2) is -(18)/(t^3)