What is the derivative of-16t^2+44t+31 ?

The derivative of-16t^2+44t+31 is -32t+44

What is the first derivative of-16t^2+44t+31 ?

The first derivative of-16t^2+44t+31 is -32t+44

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derivative of

- 16 t^{2} + 44 t + 31

- 32 t + 44

Solution steps

\frac{d}{d t} (- 16 t^{2} + 44 t + 31)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (16 t^{2}) + \frac{d}{d t} (44 t) + \frac{d}{d t} (31)

\frac{d}{d t} (16 t^{2}) = 32 t

\frac{d}{d t} (44 t) = 44

\frac{d}{d t} (31) = 0

= - 32 t + 44 + 0

Simplify

= - 32 t + 44

x \to 2 lim (3 x - 5)

x 1 - y^{2} d x = d y

\frac{d y}{d x} = y x sin (x^{2})

\int \frac{5}{5 x + 2} d x

\frac{d}{d x} (\frac{1 - ln ( x )}{x ^{2}})

What is the derivative of-16t^2+44t+31 ?
The derivative of-16t^2+44t+31 is -32t+44
What is the first derivative of-16t^2+44t+31 ?
The first derivative of-16t^2+44t+31 is -32t+44