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

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

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+y^2-yx^2-xy)

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

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

Solution steps

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

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

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

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

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

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

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

Simplify

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

\frac{d}{d t} (5 e^{t} cos (t))

f (x) = cos (5 x)

12 x^{3} - 3 y x^{4} + 1 y^{^{'}} = 0

\int \frac{e ^{u}}{u} d u

(x^{2} + 4)^{3}

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