What is the derivative of 9t^2+sqrt(t^3) ?

The derivative of 9t^2+sqrt(t^3) is 18t+(3sqrt(t))/2

What is the first derivative of 9t^2+sqrt(t^3) ?

The first derivative of 9t^2+sqrt(t^3) is 18t+(3sqrt(t))/2

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derivative of

9 t^{2} + t^{3}

18 t + \frac{3 t}{2}

Solution steps

\frac{d}{d t} (9 t^{2} + t^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (9 t^{2}) + \frac{d}{d t} (t^{3})

\frac{d}{d t} (9 t^{2}) = 18 t

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

= 18 t + \frac{3 t}{2}

\int e^{- s x} x d x

t^{2} cos (2 t)

\frac{d}{d x} (lo g_{10} (2 x - 3))

\int \frac{x ^{2} + 12 x - 4}{x ^{3} - 4 x} d x

\frac{\partial}{\partial x} (4 x ln (y) + y ln (z) - 4 x)

What is the derivative of 9t^2+sqrt(t^3) ?
The derivative of 9t^2+sqrt(t^3) is 18t+(3sqrt(t))/2
What is the first derivative of 9t^2+sqrt(t^3) ?
The first derivative of 9t^2+sqrt(t^3) is 18t+(3sqrt(t))/2