What is the derivative of sqrt(r)+\sqrt[5]{r} ?

The derivative of sqrt(r)+\sqrt[5]{r} is 1/(2sqrt(r))+1/(5r^{4/5)}

What is the first derivative of sqrt(r)+\sqrt[5]{r} ?

The first derivative of sqrt(r)+\sqrt[5]{r} is 1/(2sqrt(r))+1/(5r^{4/5)}

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}{5 r ^{\frac{4}{5}}}

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}{5 r ^{\frac{4}{5}}}

\int 1 + x^{- \frac{2}{3}} d x

\frac{\partial}{\partial x} (\frac{1}{x + 4 y})

\int 28 t (7 t^{2} - 5)^{3} d t

\frac{\partial}{\partial y} (\frac{2 x}{x ^{2} + y ^{2}})

\int \frac{x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

What is the derivative of sqrt(r)+\sqrt[5]{r} ?
The derivative of sqrt(r)+\sqrt[5]{r} is 1/(2sqrt(r))+1/(5r^{4/5)}
What is the first derivative of sqrt(r)+\sqrt[5]{r} ?
The first derivative of sqrt(r)+\sqrt[5]{r} is 1/(2sqrt(r))+1/(5r^{4/5)}