What is the d/(dt)(\sqrt[3]{t}(t^2+4)) ?

The d/(dt)(\sqrt[3]{t}(t^2+4)) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}

What is the first d/(dt)(\sqrt[3]{t}(t^2+4)) ?

The first d/(dt)(\sqrt[3]{t}(t^2+4)) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}

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Solution steps

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

\frac{d}{d t} (t^{2} + 4) = 2 t

Simplify

\int - \frac{4}{x ^{2}} d x

\frac{d y}{d x} = \frac{y}{y ^{2}}, y (0) = 2

n = 1 \sum \infty \frac{1}{2 + sin ( n )}

\frac{d}{d x} (e^{14 x})

What is the d/(dt)(\sqrt[3]{t}(t^2+4)) ?
The d/(dt)(\sqrt[3]{t}(t^2+4)) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}
What is the first d/(dt)(\sqrt[3]{t}(t^2+4)) ?
The first d/(dt)(\sqrt[3]{t}(t^2+4)) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}