What is the derivative of y=\sqrt[3]{t}(t^2+4) ?

The derivative of y=\sqrt[3]{t}(t^2+4) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}

What is the first derivative of y=\sqrt[3]{t}(t^2+4) ?

The first derivative of y=\sqrt[3]{t}(t^2+4) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}

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derivative of

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

y^{^{'}} = \frac{( x y ^{3} )}{25}

\frac{d y}{d x} = \frac{6 y}{x ^{2}}

\int - 18 t + 4 d t

\int csc^{2} (x) + 5 d x

(1 - x)^{^{'}}

What is the derivative of y=\sqrt[3]{t}(t^2+4) ?
The derivative of y=\sqrt[3]{t}(t^2+4) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}
What is the first derivative of y=\sqrt[3]{t}(t^2+4) ?
The first derivative of y=\sqrt[3]{t}(t^2+4) is (t^2+4+6\sqrt[3]{t}t^{5/3})/(3t^{2/3)}