What is the derivative of (\sqrt[3]{t})/(t-3) ?

The derivative of (\sqrt[3]{t})/(t-3) is (t-3-3\sqrt[3]{t}t^{2/3})/(3t^{2/3)(t-3)^2}

What is the first derivative of (\sqrt[3]{t})/(t-3) ?

The first derivative of (\sqrt[3]{t})/(t-3) is (t-3-3\sqrt[3]{t}t^{2/3})/(3t^{2/3)(t-3)^2}

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derivative of

Solution steps

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

\frac{d}{d t} (t - 3) = 1

Simplify

\frac{d}{d x} ((f (x) (6 x))^{2} - 4)

\frac{\partial}{\partial x} (x^{3} - 5 x^{2} y + 7 x y^{3})

\int \frac{1}{2 - e ^{- x}} d x

\frac{\partial}{\partial y} (x^{2} - x yz)

\int e^{3 x} - \frac{1}{( x + 4 ) ^{2}} d x

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