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

The derivative of x/(\sqrt[3]{x^2-1}) is (3\sqrt[3]{x^2-1}(x^2-1)^{2/3}-2x^2)/(3(x^2-1)^{4/3)}

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

The first derivative of x/(\sqrt[3]{x^2-1}) is (3\sqrt[3]{x^2-1}(x^2-1)^{2/3}-2x^2)/(3(x^2-1)^{4/3)}

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Solution steps

Apply the Quotient Rule:

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

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

Simplify

\frac{d}{d x} (2 e^{x} (cos (x) - sin (x)))

f (x) = (x - x^{2})

(y) d x + (y - x) d y = 0

x y^{^{'}} - 3 y + 12 = 0

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