What is the derivative of-2t^2-8t+9 ?

The derivative of-2t^2-8t+9 is -4t-8

What is the first derivative of-2t^2-8t+9 ?

The first derivative of-2t^2-8t+9 is -4t-8

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derivative of

- 2 t^{2} - 8 t + 9

- 4 t - 8

Solution steps

\frac{d}{d t} (- 2 t^{2} - 8 t + 9)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (2 t^{2}) - \frac{d}{d t} (8 t) + \frac{d}{d t} (9)

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

\frac{d}{d t} (8 t) = 8

\frac{d}{d t} (9) = 0

= - 4 t - 8 + 0

Simplify

= - 4 t - 8

x \to 6 - lim (\frac{x - 6}{∣ x - 6 ∣})

\frac{d}{d x} (ln (x^{x}))

\int 4 x e^{- 2 x^{2}} d x

\frac{d}{d x} ((cos (x) + 2)^{3})

\int \frac{4 x}{( x ^{2} + 6 ) ^{2}} d x

What is the derivative of-2t^2-8t+9 ?
The derivative of-2t^2-8t+9 is -4t-8
What is the first derivative of-2t^2-8t+9 ?
The first derivative of-2t^2-8t+9 is -4t-8