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

The derivative of (x^4/8+1/(4y^2)) is (x^3)/2-(\frac{dy)/(dx)}{2y^3}

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

The first derivative of (x^4/8+1/(4y^2)) is (x^3)/2-(\frac{dy)/(dx)}{2y^3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\frac{x ^{3}}{2} - \frac{\frac{d y}{d x}}{2 y ^{3}}

Solution steps

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

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

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

= \frac{x ^{3}}{2} - \frac{\frac{d y}{d x}}{2 y ^{3}}

What is the derivative of (x^4/8+1/(4y^2)) ?
The derivative of (x^4/8+1/(4y^2)) is (x^3)/2-(\frac{dy)/(dx)}{2y^3}
What is the first derivative of (x^4/8+1/(4y^2)) ?
The first derivative of (x^4/8+1/(4y^2)) is (x^3)/2-(\frac{dy)/(dx)}{2y^3}