What is the derivative of g(t)=t-4/(t^3) ?

The derivative of g(t)=t-4/(t^3) is 1+(12)/(t^4)

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

The first derivative of g(t)=t-4/(t^3) is 1+(12)/(t^4)

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derivative of

g (t) = t - \frac{4}{t ^{3}}

1 + \frac{12}{t ^{4}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d t}{d t} = 1

\frac{d}{d t} (\frac{4}{t ^{3}}) = - \frac{12}{t ^{4}}

= 1 - (- \frac{12}{t ^{4}})

Simplify

= 1 + \frac{12}{t ^{4}}

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What is the derivative of g(t)=t-4/(t^3) ?
The derivative of g(t)=t-4/(t^3) is 1+(12)/(t^4)
What is the first derivative of g(t)=t-4/(t^3) ?
The first derivative of g(t)=t-4/(t^3) is 1+(12)/(t^4)