What is the derivative of f(t)=3t+4(t)^2 ?

The derivative of f(t)=3t+4(t)^2 is 3+8t

What is the first derivative of f(t)=3t+4(t)^2 ?

The first derivative of f(t)=3t+4(t)^2 is 3+8t

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derivative of

f (t) = 3 t + 4 (t)^{2}

3 + 8 t

Solution steps

\frac{d}{d t} (3 t + 4 t^{2})

Apply the Sum/Difference Rule:

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

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

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

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

= 3 + 8 t

\int \frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{2 x ^{2}} d x

\frac{d}{d x} (x cos (x) sin (x))

\int \frac{t}{t ^{4} + 7} d t

- x cos (x)

d y = (x + 2 y) d x

What is the derivative of f(t)=3t+4(t)^2 ?
The derivative of f(t)=3t+4(t)^2 is 3+8t
What is the first derivative of f(t)=3t+4(t)^2 ?
The first derivative of f(t)=3t+4(t)^2 is 3+8t