What is the derivative of G(r)=sqrt(r)+\sqrt[4]{r} ?

The derivative of G(r)=sqrt(r)+\sqrt[4]{r} is 1/(2sqrt(r))+1/(4r^{3/4)}

What is the first derivative of G(r)=sqrt(r)+\sqrt[4]{r} ?

The first derivative of G(r)=sqrt(r)+\sqrt[4]{r} is 1/(2sqrt(r))+1/(4r^{3/4)}

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derivative of

\frac{1}{2 r} + \frac{1}{4 r ^{\frac{3}{4}}}

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}{4 r ^{\frac{3}{4}}}

What is the derivative of G(r)=sqrt(r)+\sqrt[4]{r} ?
The derivative of G(r)=sqrt(r)+\sqrt[4]{r} is 1/(2sqrt(r))+1/(4r^{3/4)}
What is the first derivative of G(r)=sqrt(r)+\sqrt[4]{r} ?
The first derivative of G(r)=sqrt(r)+\sqrt[4]{r} is 1/(2sqrt(r))+1/(4r^{3/4)}