What is (\partial)/(\partial y)(xyz+sin(2xy)) ?

The answer to (\partial)/(\partial y)(xyz+sin(2xy)) is xz+cos(2xy)*2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial y)(xyz+sin(2xy))

\frac{\partial}{\partial y} (x yz + sin (2 x y))

x z + cos (2 x y) \cdot 2 x

Solution steps

\frac{\partial}{\partial y} (x yz + sin (2 x y))

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x yz) + \frac{\partial}{\partial y} (sin (2 x y))

\frac{\partial}{\partial y} (x yz) = x z

\frac{\partial}{\partial y} (sin (2 x y)) = cos (2 x y) \cdot 2 x

= x z + cos (2 x y) \cdot 2 x

What is (\partial)/(\partial y)(xyz+sin(2xy)) ?
The answer to (\partial)/(\partial y)(xyz+sin(2xy)) is xz+cos(2xy)*2x