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

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

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

2 y x - y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y x - y^{2}

e^{(x - 1)^{2}}

cot (x^{8})

y = (\frac{8 x ^{2} - 5}{2 - x})^{3}

g (x) = lo g_{7} (8 x^{3} - 9 x^{2})

\int x (x sin (x) - cos (x)) d x

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