What is (\partial)/(\partial y)(4x^2y^3-1) ?

The answer to (\partial)/(\partial y)(4x^2y^3-1) is 12x^2y^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial y)(4x^2y^3-1)

\frac{\partial}{\partial y} (4 x^{2} y^{3} - 1)

12 x^{2} y^{2}

Solution steps

\frac{\partial}{\partial y} (4 x^{2} y^{3} - 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 x^{2} y^{3}) - \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (4 x^{2} y^{3}) = 12 x^{2} y^{2}

\frac{\partial}{\partial y} (1) = 0

= 12 x^{2} y^{2} - 0

Simplify

= 12 x^{2} y^{2}

What is (\partial)/(\partial y)(4x^2y^3-1) ?
The answer to (\partial)/(\partial y)(4x^2y^3-1) is 12x^2y^2