What is the derivative of 1/3 t^3-4t^2+16t+7 ?

The derivative of 1/3 t^3-4t^2+16t+7 is t^2-8t+16

What is the first derivative of 1/3 t^3-4t^2+16t+7 ?

The first derivative of 1/3 t^3-4t^2+16t+7 is t^2-8t+16

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derivative of

\frac{1}{3} t^{3} - 4 t^{2} + 16 t + 7

t^{2} - 8 t + 16

Solution steps

\frac{d}{d t} (\frac{1}{3} t^{3} - 4 t^{2} + 16 t + 7)

Simplify

\frac{1}{3} t^{3} - 4 t^{2} + 16 t + 7 : \frac{t ^{3}}{3} - 4 t^{2} + 16 t + 7

= \frac{d}{d t} (\frac{t ^{3}}{3} - 4 t^{2} + 16 t + 7)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{t ^{3}}{3}) - \frac{d}{d t} (4 t^{2}) + \frac{d}{d t} (16 t) + \frac{d}{d t} (7)

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

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

\frac{d}{d t} (16 t) = 16

\frac{d}{d t} (7) = 0

= t^{2} - 8 t + 16 + 0

Simplify

= t^{2} - 8 t + 16

x \to 0 lim (lo g_{2} (x))

\int (x + 6 y) d y

\int (2 x) d x

x \to 0 lim (9^{x})

\frac{s}{s ^{2} + 2 s - 3}

What is the derivative of 1/3 t^3-4t^2+16t+7 ?
The derivative of 1/3 t^3-4t^2+16t+7 is t^2-8t+16
What is the first derivative of 1/3 t^3-4t^2+16t+7 ?
The first derivative of 1/3 t^3-4t^2+16t+7 is t^2-8t+16