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

The answer to (\partial)/(\partial x)(x^3y+xy^2-2x^2-y^2) is 3x^2y-4x+y^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

3 x^{2} y - 4 x + y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 3 y x^{2} + y^{2} - 4 x - 0

Simplify

= 3 x^{2} y - 4 x + y^{2}

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

\int u \cdot sec^{2} (u) d u

f (x) = 8 x^{3}

\frac{d}{d x} ((sec^{2} (x)) d x)

f (t) = t^{6} sin (t)

What is (\partial)/(\partial x)(x^3y+xy^2-2x^2-y^2) ?
The answer to (\partial)/(\partial x)(x^3y+xy^2-2x^2-y^2) is 3x^2y-4x+y^2