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

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

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

The first derivative of y=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{d}{d x} (4 x^{4} + \frac{1}{3} x^{3} + x + 1)

h

\frac{d}{d x} (\frac{1 - sin ( x )}{1 + sin ( x )})

\int e^{e x} d x

\int (x^{2} + 1 + \frac{1}{x ^{2} + 1}) d x

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