What is the derivative of f(t)=t^2+4t ?

The derivative of f(t)=t^2+4t is 2t+4

What is the first derivative of f(t)=t^2+4t ?

The first derivative of f(t)=t^2+4t is 2t+4

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derivative of

f (t) = t^{2} + 4 t

2 t + 4

Solution steps

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

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} (t^{2}) = 2 t

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

= 2 t + 4

\int \frac{1}{( 1 + y ) ^{2}} d y

\int_{- 1}^{1} 5^{x} d x

x \to 0 - lim (\frac{1}{x ^{2}})

\int \frac{( ln ( x ) ) ^{6}}{x} d x

l a pl a ce t r an s f or m t \cdot e^{- 4 t} cos (5 t)

What is the derivative of f(t)=t^2+4t ?
The derivative of f(t)=t^2+4t is 2t+4
What is the first derivative of f(t)=t^2+4t ?
The first derivative of f(t)=t^2+4t is 2t+4