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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

3 x^{2} - 3 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial x} (2) = 0

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

Simplify

= 3 x^{2} - 3 y

\frac{4 s + 1}{2 s ^{2} - 5 s + 6}

f (x) = π^{x^{2}}

y^{^{'}} = 2

\int 8 x^{- 5} d x

\int \frac{1}{x ^{2} + 100} d x

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