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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial z} (e^{x} cos (y)) = 0

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

= 0 + y

Simplify

= y

n = 1 \sum \infty 2^{- 4 n}

\int sin (2 π t) d t

\int \frac{1}{y + y} d y

\int \frac{x ^{3}}{x ^{2} + 30 x + 225} d x

t \to 2 lim (\frac{2 t}{4 - t ^{2}})

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