What is the derivative of y=\sqrt[3]{t}+2sqrt(t^3) ?

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

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

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

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derivative of

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

Solution steps

Apply the Sum/Difference Rule:

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

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

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

\int_{0}^{3} \frac{1}{( x - 1 ) ^{2}} d x

y = 6 x - x^{2}, y = 2 x

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

\frac{d y}{d x} = 2.5 - \frac{5 y}{2 x + 100}

\int (9 - x^{2})^{2} d x

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