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

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

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

The first derivative of f(x)=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}}}

\frac{e ^{- π s}}{s - 1}

\frac{d}{d x} (2 x^{6})

\frac{\partial}{\partial x} ((7 x + 2) (9 x - 4))

y = x^{3} - x^{2} - 12 x, x = - 3, x = 0

\frac{d}{d x} (\frac{- sin ^{3} ( ln ( x ) )}{x})

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