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

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

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

The first derivative of G(r)=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}}}

\int x x^{2} d x

x \to 2 lim (2 x^{2} + 5 x + 3)

\int 2 x (x^{2} + 6)^{3} d x

\int \frac{sin ^{2} ( \frac{1}{t} )}{t ^{2}} d t

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