What is the derivative of \sqrt[3]{t}+4sqrt(t^3) ?

The derivative of \sqrt[3]{t}+4sqrt(t^3) is 1/(3t^{2/3)}+6sqrt(t)

What is the first derivative of \sqrt[3]{t}+4sqrt(t^3) ?

The first derivative of \sqrt[3]{t}+4sqrt(t^3) is 1/(3t^{2/3)}+6sqrt(t)

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derivative of

\frac{1}{3 t ^{\frac{2}{3}}} + 6 t

Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{1}{3 t ^{\frac{2}{3}}} + 6 t

x \to - 0.001 lim (4 x^{2} + 7)

\frac{\partial}{\partial x} (4 x^{3} y^{3} - 18 y - x)

\frac{\partial}{\partial y} (9 x sin (y - z))

\int_{- 4}^{- 1} x^{4} d x

\int 20 tan^{6} (x) d x

What is the derivative of \sqrt[3]{t}+4sqrt(t^3) ?
The derivative of \sqrt[3]{t}+4sqrt(t^3) is 1/(3t^{2/3)}+6sqrt(t)
What is the first derivative of \sqrt[3]{t}+4sqrt(t^3) ?
The first derivative of \sqrt[3]{t}+4sqrt(t^3) is 1/(3t^{2/3)}+6sqrt(t)