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

The derivative of g(t)=t^2-2/(t^3) is 2t+6/(t^4)

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

The first derivative of g(t)=t^2-2/(t^3) is 2t+6/(t^4)

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derivative of

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

2 t + \frac{6}{t ^{4}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d t} (\frac{2}{t ^{3}}) = - \frac{6}{t ^{4}}

= 2 t - (- \frac{6}{t ^{4}})

Simplify

= 2 t + \frac{6}{t ^{4}}

\int t^{2} e^{- 20 t} d t

\int_{t}^{1} ln (x) d x

\frac{1}{s ^{2} + 2}

\int \frac{x ^{2}}{x ^{2} - 2 x + 1} d x

\frac{1}{8 x ^{2}}

What is the derivative of g(t)=t^2-2/(t^3) ?
The derivative of g(t)=t^2-2/(t^3) is 2t+6/(t^4)
What is the first derivative of g(t)=t^2-2/(t^3) ?
The first derivative of g(t)=t^2-2/(t^3) is 2t+6/(t^4)