What is the derivative of G(r)=sqrt(r)+\sqrt[6]{r} ?

The derivative of G(r)=sqrt(r)+\sqrt[6]{r} is 1/(2sqrt(r))+1/(6r^{5/6)}

What is the first derivative of G(r)=sqrt(r)+\sqrt[6]{r} ?

The first derivative of G(r)=sqrt(r)+\sqrt[6]{r} is 1/(2sqrt(r))+1/(6r^{5/6)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{1}{2 r} + \frac{1}{6 r ^{\frac{5}{6}}}

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}{6 r ^{\frac{5}{6}}}

x \frac{d y}{d x} = y - x^{2} + y^{2}

\frac{d}{d x} (\frac{10}{2 x})

\frac{d}{d x} (\frac{x}{2} ln (\frac{1 + x}{1 - x}) - 1)

8 sin (2 x)

\frac{d}{d x} (\frac{1 + x}{x ^{4}})

What is the derivative of G(r)=sqrt(r)+\sqrt[6]{r} ?
The derivative of G(r)=sqrt(r)+\sqrt[6]{r} is 1/(2sqrt(r))+1/(6r^{5/6)}
What is the first derivative of G(r)=sqrt(r)+\sqrt[6]{r} ?
The first derivative of G(r)=sqrt(r)+\sqrt[6]{r} is 1/(2sqrt(r))+1/(6r^{5/6)}