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

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

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

The first derivative of y=\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

x \to + 0 lim (\frac{- 6}{x ^{7}})

\int_{0}^{3} ∣ x - 2 ∣ d x

\int \frac{1}{x ^{2} - 2 x + 65} d x

f (x) = x^{2} + 3 x, at x = - 2

\frac{( z - 1 )}{z + 1}

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