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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- 4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 4 y

Simplify

= - 4 y

\frac{d}{d x} (\frac{x}{a})

t an g e n t f (x) = (1 + 4 x) (5 - 2 x), at x = 2

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

t an g e n t f (x) = sin (sin (x)), 0, at x = π

\int \frac{( x ^{2} )}{x ^{2} - 1} d x

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