What is the derivative of s(t)=14t^2+840t ?

The derivative of s(t)=14t^2+840t is 28t+840

What is the first derivative of s(t)=14t^2+840t ?

The first derivative of s(t)=14t^2+840t is 28t+840

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derivative of

s (t) = 14 t^{2} + 840 t

28 t + 840

Solution steps

\frac{d}{d t} (14 t^{2} + 840 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (14 t^{2}) + \frac{d}{d t} (840 t)

\frac{d}{d t} (14 t^{2}) = 28 t

\frac{d}{d t} (840 t) = 840

= 28 t + 840

0

x \to 0 lim (\frac{0 \cdot x ^{6}}{x ^{9}})

3 y^{^{''}} + 75 y = 0

f (x) = 4 - \frac{20}{x ^{3}} + \frac{3}{x ^{5}}

\frac{d}{d x} (lo g_{6} (x^{2} - 3 x))

What is the derivative of s(t)=14t^2+840t ?
The derivative of s(t)=14t^2+840t is 28t+840
What is the first derivative of s(t)=14t^2+840t ?
The first derivative of s(t)=14t^2+840t is 28t+840