What is the derivative of 9/(t^2)-3/t ?

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

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

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

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derivative of

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

- \frac{18}{t ^{3}} + \frac{3}{t ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (\frac{9}{t ^{2}}) - \frac{d}{d t} (\frac{3}{t})

\frac{d}{d t} (\frac{9}{t ^{2}}) = - \frac{18}{t ^{3}}

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

= - \frac{18}{t ^{3}} - (- \frac{3}{t ^{2}})

Simplify

= - \frac{18}{t ^{3}} + \frac{3}{t ^{2}}

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