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

The derivative of g(t)=-4t^2+3t-3 is -8t+3

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

The first derivative of g(t)=-4t^2+3t-3 is -8t+3

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derivative of

g (t) = - 4 t^{2} + 3 t - 3

- 8 t + 3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d t} (4 t^{2}) = 8 t

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

\frac{d}{d t} (3) = 0

= - 8 t + 3 - 0

Simplify

= - 8 t + 3

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}

\int_{- 1}^{4} (x^{2} + x^{3}) d x

x^{3}

f (x) = \frac{1 - 3 ( a - 1 )}{8}

\frac{d}{d x} (\frac{2 x}{4 x ^{2} + 1})

What is the derivative of g(t)=-4t^2+3t-3 ?
The derivative of g(t)=-4t^2+3t-3 is -8t+3
What is the first derivative of g(t)=-4t^2+3t-3 ?
The first derivative of g(t)=-4t^2+3t-3 is -8t+3