What is (\partial)/(\partial y)(cos(x+y+z)) ?

The answer to (\partial)/(\partial y)(cos(x+y+z)) is -sin(x+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)(cos(x+y+z))

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

- sin (x + y + z)

Solution steps

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

Treat

x, z

as constants

Apply the chain rule:

- sin (x + y + z) \frac{\partial}{\partial y} (x + y + z)

= - sin (x + y + z) \frac{\partial}{\partial y} (x + y + z)

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

= - sin (x + y + z) \cdot 1

Simplify

= - sin (x + y + z)

x \to \infty lim (\frac{x}{6})

x \to 1 lim (ln (1 - x))

n = 1 \sum \infty \frac{1}{2 ^{n} \cdot n !}

\frac{d}{d x} (\frac{e ^{2 x}}{1 + e ^{2 x}})

5 x^{2} - y^{2} = 3

What is (\partial)/(\partial y)(cos(x+y+z)) ?
The answer to (\partial)/(\partial y)(cos(x+y+z)) is -sin(x+y+z)