What is the derivative of (2x/(\sqrt[3]{x)-2}) ?

The derivative of (2x/(\sqrt[3]{x)-2}) is (2(3\sqrt[3]{x}-6-x^{1/3}))/(3(\sqrt[3]{x)-2)^2}

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

The first derivative of (2x/(\sqrt[3]{x)-2}) is (2(3\sqrt[3]{x}-6-x^{1/3}))/(3(\sqrt[3]{x)-2)^2}

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Solution steps

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply the Quotient Rule:

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

\frac{d x}{d x} = 1

Simplify

\int_{1}^{2} 5 x d x

f (x) = 2^{x} 1

\frac{d y}{d t} - 2 t y = 3 t^{2} e^{t^{2}}

\int_{0}^{π} cos^{2} (t) d t

f (x) = x^{e^{x}}

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