What is (\partial)/(\partial y)(e^xcos(y)+yz) ?

The answer to (\partial)/(\partial y)(e^xcos(y)+yz) is -e^xsin(y)+z

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(e^xcos(y)+yz)

\frac{\partial}{\partial y} (e^{x} cos (y) + yz)

- e^{x} sin (y) + z

Solution steps

\frac{\partial}{\partial y} (e^{x} cos (y) + yz)

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (e^{x} cos (y)) + \frac{\partial}{\partial y} (yz)

\frac{\partial}{\partial y} (e^{x} cos (y)) = - e^{x} sin (y)

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

= - e^{x} sin (y) + z

30 x - 3 x

\frac{d}{d x} (e^{- (x^{2} + y^{2})})

y sin (8 x) = x cos (2 y), (\frac{π}{2}, \frac{π}{4})

\frac{d}{d x} (5 sec (2 x))

\int y ln (y) d y

What is (\partial)/(\partial y)(e^xcos(y)+yz) ?
The answer to (\partial)/(\partial y)(e^xcos(y)+yz) is -e^xsin(y)+z