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

The derivative of sqrt(x)+\sqrt[5]{x} is 1/(2sqrt(x))+1/(5x^{4/5)}

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

The first derivative of sqrt(x)+\sqrt[5]{x} is 1/(2sqrt(x))+1/(5x^{4/5)}

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derivative of

\frac{1}{2 x} + \frac{1}{5 x ^{\frac{4}{5}}}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{1}{2 x} + \frac{1}{5 x ^{\frac{4}{5}}}

\int \frac{1}{2 e ^{2 x}} d x

\int_{0}^{1} π (x - x^{3})^{2} d x

2 x^{3} - 6 x^{2} - 18 - 3

y = - 2 x^{2} + 6, y = - 2

What is the derivative of sqrt(x)+\sqrt[5]{x} ?
The derivative of sqrt(x)+\sqrt[5]{x} is 1/(2sqrt(x))+1/(5x^{4/5)}
What is the first derivative of sqrt(x)+\sqrt[5]{x} ?
The first derivative of sqrt(x)+\sqrt[5]{x} is 1/(2sqrt(x))+1/(5x^{4/5)}