What is the derivative of g(t)=t^3(2t^2-t^{-1}) ?

The derivative of g(t)=t^3(2t^2-t^{-1}) is 10t^4-2t

What is the first derivative of g(t)=t^3(2t^2-t^{-1}) ?

The first derivative of g(t)=t^3(2t^2-t^{-1}) is 10t^4-2t

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derivative of

g (t) = t^{3} (2 t^{2} - t^{- 1})

10 t^{4} - 2 t

Solution steps

\frac{d}{d t} (t^{3} (2 t^{2} - t^{- 1}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{3}, g = 2 t^{2} - t^{- 1}

= \frac{d}{d t} (t^{3}) (2 t^{2} - t^{- 1}) + \frac{d}{d t} (2 t^{2} - t^{- 1}) t^{3}

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

\frac{d}{d t} (2 t^{2} - t^{- 1}) = 4 t + \frac{1}{t ^{2}}

= 3 t^{2} (2 t^{2} - t^{- 1}) + (4 t + \frac{1}{t ^{2}}) t^{3}

Simplify

3 t^{2} (2 t^{2} - t^{- 1}) + (4 t + \frac{1}{t ^{2}}) t^{3} : 10 t^{4} - 2 t

= 10 t^{4} - 2 t

3 x^{2}, 3 x^{3}

y = 1 + 2 x - x^{3} (1.2)

\frac{d}{d x} (x^{2} ∣ x ∣)

\frac{d}{d x} (\frac{x ^{2}}{cos ( x )})

\frac{d}{d x} (3 x^{2} - \frac{3}{x ^{4}})

What is the derivative of g(t)=t^3(2t^2-t^{-1}) ?
The derivative of g(t)=t^3(2t^2-t^{-1}) is 10t^4-2t
What is the first derivative of g(t)=t^3(2t^2-t^{-1}) ?
The first derivative of g(t)=t^3(2t^2-t^{-1}) is 10t^4-2t