What is the derivative of (\sqrt[5]{x^2}+\sqrt[4]{x^5})/(2x) ?

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

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

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

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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 x}{d x} = 1

Simplify

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