What is the derivative of sqrt(r)+\sqrt[8]{r} ?

The derivative of sqrt(r)+\sqrt[8]{r} is 1/(2sqrt(r))+1/(8r^{7/8)}

What is the first derivative of sqrt(r)+\sqrt[8]{r} ?

The first derivative of sqrt(r)+\sqrt[8]{r} is 1/(2sqrt(r))+1/(8r^{7/8)}

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derivative of

\frac{1}{2 r} + \frac{1}{8 r ^{\frac{7}{8}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{1}{2 r} + \frac{1}{8 r ^{\frac{7}{8}}}

\frac{d}{d x} (\frac{1 + x}{x ^{3}})

\frac{\partial}{\partial x} (x y e^{y})

y^{^{'}} + 7 y = e^{6 x}

\int_{0}^{2 π} sin^{2} (x) d x

n = 1 \sum \infty n^{- \frac{1}{5}}

What is the derivative of sqrt(r)+\sqrt[8]{r} ?
The derivative of sqrt(r)+\sqrt[8]{r} is 1/(2sqrt(r))+1/(8r^{7/8)}
What is the first derivative of sqrt(r)+\sqrt[8]{r} ?
The first derivative of sqrt(r)+\sqrt[8]{r} is 1/(2sqrt(r))+1/(8r^{7/8)}