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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

12 x^{2} y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (4) = 0

= 12 x^{2} y^{3} - 0

Simplify

= 12 x^{2} y^{3}

\frac{2}{x ^{2} + 3}

2 a x + b

\frac{d y}{d x} = \frac{2 x}{y}

\int_{- 8}^{8} x^{\frac{1}{3}} d x

y^{^{''}} - \frac{1}{x} y^{^{'}} = 3 x

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