What is (\partial)/(\partial y)(x^2y+y^2z+z^2x) ?

The answer to (\partial)/(\partial y)(x^2y+y^2z+z^2x) is x^2+2yz

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(x^2y+y^2z+z^2x)

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

x^{2} + 2 yz

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (y^{2} z) = 2 zy

\frac{\partial}{\partial y} (z^{2} x) = 0

= x^{2} + 2 zy + 0

Simplify

= x^{2} + 2 yz

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 5}{2 x - 1})

y^{^{'}} + x y = x y^{- 2}

\frac{d}{d x} (12 x - 3 x^{2})

\frac{1}{1 + 3 x}

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

What is (\partial)/(\partial y)(x^2y+y^2z+z^2x) ?
The answer to (\partial)/(\partial y)(x^2y+y^2z+z^2x) is x^2+2yz