What is (\partial}{\partial y}((x^4)/(6y)+\frac{12y^4)/x) ?

The answer to (\partial}{\partial y}((x^4)/(6y)+\frac{12y^4)/x) is -(x^4}{6y^2}+\frac{48y^3)/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial}{\partial y}((x^4)/(6y)+\frac{12y^4)/x)

\frac{\partial}{\partial y} (\frac{x ^{4}}{6 y} + \frac{12 y ^{4}}{x})

- \frac{x ^{4}}{6 y ^{2}} + \frac{48 y ^{3}}{x}

Solution steps

\frac{\partial}{\partial y} (\frac{x ^{4}}{6 y} + \frac{12 y ^{4}}{x})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (\frac{x ^{4}}{6 y}) + \frac{\partial}{\partial y} (\frac{12 y ^{4}}{x})

\frac{\partial}{\partial y} (\frac{x ^{4}}{6 y}) = - \frac{x ^{4}}{6 y ^{2}}

\frac{\partial}{\partial y} (\frac{12 y ^{4}}{x}) = \frac{48 y ^{3}}{x}

= - \frac{x ^{4}}{6 y ^{2}} + \frac{48 y ^{3}}{x}

\int 2 sin^{2} (θ) d θ

\int_{- 6}^{6} \frac{1}{2} (36 - x^{2}) d x

\int_{- π}^{π} sin (2 x) sin (3 x) d x

f (x) = e^{x^{5}}

\int \frac{10}{10 + e ^{x}} d x

What is (\partial}{\partial y}((x^4)/(6y)+\frac{12y^4)/x) ?
The answer to (\partial}{\partial y}((x^4)/(6y)+\frac{12y^4)/x) is -(x^4}{6y^2}+\frac{48y^3)/x