What is the derivative of-t^2+4t+4 ?

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

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

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

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derivative of

- t^{2} + 4 t + 4

- 2 t + 4

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d t} (4) = 0

= - 2 t + 4 + 0

Simplify

= - 2 t + 4

\frac{d}{d x} ((i) \cdot (x^{- 5})^{5})

e^{2 x - y + z}

\frac{d}{d a} (\frac{a}{18})

e^{s} in x + x (tan (x))^{2}

(4 x + 8)^{\frac{1}{3}}

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