What is (\partial)/(\partial y)(x^2+y^2+z^2-1) ?

The answer to (\partial)/(\partial y)(x^2+y^2+z^2-1) is 2y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial y)(x^2+y^2+z^2-1)

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2} - 1)

2 y

Solution steps

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2} - 1)

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{2}) + \frac{\partial}{\partial y} (y^{2}) + \frac{\partial}{\partial y} (z^{2}) - \frac{\partial}{\partial y} (1)

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

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

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

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

= 0 + 2 y + 0 - 0

Simplify

= 2 y

What is (\partial)/(\partial y)(x^2+y^2+z^2-1) ?
The answer to (\partial)/(\partial y)(x^2+y^2+z^2-1) is 2y