What is (\partial)/(\partial y)(x^3+yz^2) ?

The answer to (\partial)/(\partial y)(x^3+yz^2) is z^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(x^3+yz^2)

\frac{\partial}{\partial y} (x^{3} + y z^{2})

z^{2}

Solution steps

\frac{\partial}{\partial y} (x^{3} + y z^{2})

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{3}) + \frac{\partial}{\partial y} (y z^{2})

\frac{\partial}{\partial y} (x^{3}) = 0

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

= 0 + z^{2}

Simplify

= z^{2}

\int \frac{1}{x - 2} d x

\frac{d}{d x} (csc (cot (x)))

x \to 1 + lim (\frac{x + 2}{x - 1})

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

\int (3 x - x^{- 1})^{2} d x

What is (\partial)/(\partial y)(x^3+yz^2) ?
The answer to (\partial)/(\partial y)(x^3+yz^2) is z^2