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

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

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

The first derivative of f(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}}}

\int 17 cot^{4} (x) d x

\frac{d}{d t} (e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t))

\frac{d}{d x} (- 2 x^{5} - 3 x^{3} + 1)

x \to 2 - lim (∣ x ∣ - x)

y^{^{''}} + \frac{2 y ^{^{'}}}{x} - \frac{6 y}{x ^{2}} = 0

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