What is the derivative of \sqrt[3]{t}+6sqrt(t^3) ?

The derivative of \sqrt[3]{t}+6sqrt(t^3) is 1/(3t^{2/3)}+9sqrt(t)

What is the first derivative of \sqrt[3]{t}+6sqrt(t^3) ?

The first derivative of \sqrt[3]{t}+6sqrt(t^3) is 1/(3t^{2/3)}+9sqrt(t)

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derivative of

\frac{1}{3 t ^{\frac{2}{3}}} + 9 t

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d t} (6 t^{3}) = 9 t

= \frac{1}{3 t ^{\frac{2}{3}}} + 9 t

\int_{0}^{4 π} t^{2} + 5 d t

\frac{\partial}{\partial x} (x sin (3 y) - z e^{- 2 x})

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

x \to 5 lim (4 x - 5)

f (x) = 7 x - x^{2}

What is the derivative of \sqrt[3]{t}+6sqrt(t^3) ?
The derivative of \sqrt[3]{t}+6sqrt(t^3) is 1/(3t^{2/3)}+9sqrt(t)
What is the first derivative of \sqrt[3]{t}+6sqrt(t^3) ?
The first derivative of \sqrt[3]{t}+6sqrt(t^3) is 1/(3t^{2/3)}+9sqrt(t)