What is the derivative of (4sqrt(t^3))/3+6 ?

The derivative of (4sqrt(t^3))/3+6 is 2sqrt(t)

What is the first derivative of (4sqrt(t^3))/3+6 ?

The first derivative of (4sqrt(t^3))/3+6 is 2sqrt(t)

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derivative of

\frac{4 t ^{3}}{3} + 6

2 t

Solution steps

\frac{d}{d t} (\frac{4 t ^{3}}{3} + 6)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{4 t ^{3}}{3}) + \frac{d}{d t} (6)

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

\frac{d}{d t} (6) = 0

= 2 t + 0

Simplify

= 2 t

f (x) = (x^{2} + 1) (x^{3} - x) (3 x^{4} + 2 x - 1)

x \to 0 lim (ln (\frac{- 1}{6} x))

\int \frac{5}{3 x x ^{2} - 25} d x

\int \frac{x ^{2}}{x ^{4} - 1} d x

\int \frac{3}{x ^{3}} d x

What is the derivative of (4sqrt(t^3))/3+6 ?
The derivative of (4sqrt(t^3))/3+6 is 2sqrt(t)
What is the first derivative of (4sqrt(t^3))/3+6 ?
The first derivative of (4sqrt(t^3))/3+6 is 2sqrt(t)