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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

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

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

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

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

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

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