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

The answer to (\partial)/(\partial y)(x^3-3x+3xy^2) is 6xy

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(x^3-3x+3xy^2)

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

6 x y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (3 x y^{2}) = 6 x y

= 0 - 0 + 6 x y

Simplify

= 6 x y

f (x) = 7 - x^{2}

3 x^{2} y d x + y d y = 0

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

y = \frac{1}{x ^{3}}, (- 3, - \frac{1}{27})

\frac{d}{d x} (6 x - 7)

What is (\partial)/(\partial y)(x^3-3x+3xy^2) ?
The answer to (\partial)/(\partial y)(x^3-3x+3xy^2) is 6xy