What is the derivative of G(r)=sqrt(r)+\sqrt[9]{r} ?

The derivative of G(r)=sqrt(r)+\sqrt[9]{r} is 1/(2sqrt(r))+1/(9r^{8/9)}

What is the first derivative of G(r)=sqrt(r)+\sqrt[9]{r} ?

The first derivative of G(r)=sqrt(r)+\sqrt[9]{r} is 1/(2sqrt(r))+1/(9r^{8/9)}

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derivative of

\frac{1}{2 r} + \frac{1}{9 r ^{\frac{8}{9}}}

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}{9 r ^{\frac{8}{9}}}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 5

n \to \infty lim (n - 1)

\frac{d y}{d x} = \frac{x ^{2}}{y ( 1 + x ^{3} )}

y = \frac{4 x + 1}{x}

- \frac{5 e ^{\frac{5}{x}}}{x ^{2}}

What is the derivative of G(r)=sqrt(r)+\sqrt[9]{r} ?
The derivative of G(r)=sqrt(r)+\sqrt[9]{r} is 1/(2sqrt(r))+1/(9r^{8/9)}
What is the first derivative of G(r)=sqrt(r)+\sqrt[9]{r} ?
The first derivative of G(r)=sqrt(r)+\sqrt[9]{r} is 1/(2sqrt(r))+1/(9r^{8/9)}