What is the derivative of x^4 1/8+1/(4x^2) ?

The derivative of x^4 1/8+1/(4x^2) is (x^3)/2-1/(2x^3)

What is the first derivative of x^4 1/8+1/(4x^2) ?

The first derivative of x^4 1/8+1/(4x^2) is (x^3)/2-1/(2x^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}})

\frac{x ^{3}}{2} - \frac{1}{2 x ^{3}}

Solution steps

\frac{d}{d x} (x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}})

Simplify

x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}} : \frac{x ^{4}}{8} + \frac{1}{4 x ^{2}}

= \frac{d}{d x} (\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{4}}{8}) + \frac{d}{d x} (\frac{1}{4 x ^{2}})

\frac{d}{d x} (\frac{x ^{4}}{8}) = \frac{x ^{3}}{2}

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

= \frac{x ^{3}}{2} - \frac{1}{2 x ^{3}}

\frac{d}{d x} ((6 x + 1)^{2})

y^{^{''}} - 25 y = 0

y^{^{'}} + 27 y = \frac{1}{x ^{2}}

\int \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} d x

x^{4} + y^{3} x^{5} + 4 y^{^{'}} = 0

What is the derivative of x^4 1/8+1/(4x^2) ?
The derivative of x^4 1/8+1/(4x^2) is (x^3)/2-1/(2x^3)
What is the first derivative of x^4 1/8+1/(4x^2) ?
The first derivative of x^4 1/8+1/(4x^2) is (x^3)/2-1/(2x^3)