What is the derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) ?

The derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) is -1/(4x^{3/2)}-3/(16x^{7/4)}

What is the first derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) ?

The first derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) is -1/(4x^{3/2)}-3/(16x^{7/4)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}})

- \frac{1}{4 x ^{\frac{3}{2}}} - \frac{3}{16 x ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{2 x}) + \frac{d}{d x} (\frac{1}{4 x ^{\frac{3}{4}}})

\frac{d}{d x} (\frac{1}{2 x}) = - \frac{1}{4 x ^{\frac{3}{2}}}

\frac{d}{d x} (\frac{1}{4 x ^{\frac{3}{4}}}) = - \frac{3}{16 x ^{\frac{7}{4}}}

= - \frac{1}{4 x ^{\frac{3}{2}}} - \frac{3}{16 x ^{\frac{7}{4}}}

\frac{\partial}{\partial y} (\frac{3 x - y}{5 x y + 9})

x y^{^{'}} - y = x^{2} sin (x)

\frac{d x}{d y} = 5 x (6 - x)

\frac{\partial}{\partial x} (- \frac{2 x ^{2}}{5 y ^{2}})

x \to 0 lim (\frac{( 1 + x ) ^{5} - 1}{x})

What is the derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) ?
The derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) is -1/(4x^{3/2)}-3/(16x^{7/4)}
What is the first derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) ?
The first derivative of 1/(2sqrt(x)+1/(4x^{3/4)}) is -1/(4x^{3/2)}-3/(16x^{7/4)}