What is the derivative of 4t^2+3t+3 ?

The derivative of 4t^2+3t+3 is 8t+3

What is the first derivative of 4t^2+3t+3 ?

The first derivative of 4t^2+3t+3 is 8t+3

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derivative of

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 = x^{2} - 4 x + 1

lo g_{10} (2 - x)

\frac{d}{d x} (a \cdot e^{2 x} + b \cdot e^{- 3 x} + c \cdot e^{x})

\frac{50}{( 13 sin ( x + 33. 6 ^{\circ} ) )}

\frac{e ^{- x}}{2}

What is the derivative of 4t^2+3t+3 ?
The derivative of 4t^2+3t+3 is 8t+3
What is the first derivative of 4t^2+3t+3 ?
The first derivative of 4t^2+3t+3 is 8t+3