What is the derivative of (x+8/(\sqrt[3]{3x-4)}) ?

The derivative of (x+8/(\sqrt[3]{3x-4)}) is (\sqrt[3]{3x-4}(3x-4)^{2/3}-x-8)/((3x-4)^{4/3)}

What is the first derivative of (x+8/(\sqrt[3]{3x-4)}) ?

The first derivative of (x+8/(\sqrt[3]{3x-4)}) is (\sqrt[3]{3x-4}(3x-4)^{2/3}-x-8)/((3x-4)^{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}{d x} (x + 8) = 1

Simplify

\int_{0}^{π} x + x cos (x) d x

x \to 5 lim (\frac{4}{x})

\frac{d}{d x} (- \frac{5}{4 x ^{3}})

\frac{d}{d y} (\frac{y ^{2}}{2})

\frac{\partial}{\partial x} (cos (x^{2} + y^{2}))

What is the derivative of (x+8/(\sqrt[3]{3x-4)}) ?
The derivative of (x+8/(\sqrt[3]{3x-4)}) is (\sqrt[3]{3x-4}(3x-4)^{2/3}-x-8)/((3x-4)^{4/3)}
What is the first derivative of (x+8/(\sqrt[3]{3x-4)}) ?
The first derivative of (x+8/(\sqrt[3]{3x-4)}) is (\sqrt[3]{3x-4}(3x-4)^{2/3}-x-8)/((3x-4)^{4/3)}