What is (\partial)/(\partial y)(x+cos(xyz)-1) ?

The answer to (\partial)/(\partial y)(x+cos(xyz)-1) is -xzsin(xyz)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(x+cos(xyz)-1)

\frac{\partial}{\partial y} (x + cos (x yz) - 1)

- x z sin (x yz)

Solution steps

\frac{\partial}{\partial y} (x + cos (x yz) - 1)

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x) + \frac{\partial}{\partial y} (cos (x yz)) - \frac{\partial}{\partial y} (1)

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

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

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

= 0 - sin (x yz) x z - 0

Simplify

= - x z sin (x yz)

2 x, x^{2 - 3}

\frac{d}{d x} (sin (x) + \frac{2}{9} cot (x))

ln (14 x)

\frac{d}{d x} (\frac{108}{x})

f (x) = (2 t - 1) (6 t - 6)^{- 1}

What is (\partial)/(\partial y)(x+cos(xyz)-1) ?
The answer to (\partial)/(\partial y)(x+cos(xyz)-1) is -xzsin(xyz)