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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(3x^2+y^2)

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (3 x^{2}) = 0

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

= 0 + 2 y

Simplify

= 2 y

3 x y^{^{'}} + 4 y = \frac{x}{y}

1 + 3 t + 3 t^{2}

\frac{d}{d x} (\frac{sin ( x + 1 )}{e ^{x}})

\frac{e ^{- 4 x}}{( x - 2 ) ^{2}}

x \to \infty lim ((1 + \frac{c}{x})^{2 x})

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