What is the derivative of (6\sqrt[3]{x})/(x^2-4) ?

The derivative of (6\sqrt[3]{x})/(x^2-4) is (2(x^2-6\sqrt[3]{x}x^{5/3}-4))/(x^{2/3)(x^2-4)^2}

What is the first derivative of (6\sqrt[3]{x})/(x^2-4) ?

The first derivative of (6\sqrt[3]{x})/(x^2-4) is (2(x^2-6\sqrt[3]{x}x^{5/3}-4))/(x^{2/3)(x^2-4)^2}

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derivative of

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}{d x} (x^{2} - 4) = 2 x

Simplify

y^{^{''}} + 9 y = 5 csc^{2} (3 t)

x \to 0 lim (\frac{1}{2 + 3 ^{\frac{1}{x}}})

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

n = 0 \sum \infty (- \frac{1}{2})^{n}

What is the derivative of (6\sqrt[3]{x})/(x^2-4) ?
The derivative of (6\sqrt[3]{x})/(x^2-4) is (2(x^2-6\sqrt[3]{x}x^{5/3}-4))/(x^{2/3)(x^2-4)^2}
What is the first derivative of (6\sqrt[3]{x})/(x^2-4) ?
The first derivative of (6\sqrt[3]{x})/(x^2-4) is (2(x^2-6\sqrt[3]{x}x^{5/3}-4))/(x^{2/3)(x^2-4)^2}