What is (\partial)/(\partial z)((4x-4z)/(y+4z)) ?

The answer to (\partial)/(\partial z)((4x-4z)/(y+4z)) is (-16x-4y)/((y+4z)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial z)((4x-4z)/(y+4z))

\frac{\partial}{\partial z} (\frac{4 x - 4 z}{y + 4 z})

\frac{- 16 x - 4 y}{( y + 4 z ) ^{2}}

Solution steps

\frac{\partial}{\partial z} (\frac{4 x - 4 z}{y + 4 z})

Treat

x, y

as constants

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{\partial}{\partial z} ( 4 x - 4 z ) ( y + 4 z ) - \frac{\partial}{\partial z} ( y + 4 z ) ( 4 x - 4 z )}{( y + 4 z ) ^{2}}

\frac{\partial}{\partial z} (4 x - 4 z) = - 4

\frac{\partial}{\partial z} (y + 4 z) = 4

= \frac{( - 4 ) ( y + 4 z ) - 4 ( 4 x - 4 z )}{( y + 4 z ) ^{2}}

Simplify

\frac{( - 4 ) ( y + 4 z ) - 4 ( 4 x - 4 z )}{( y + 4 z ) ^{2}} : \frac{- 16 x - 4 y}{( y + 4 z ) ^{2}}

= \frac{- 16 x - 4 y}{( y + 4 z ) ^{2}}

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

\frac{d}{d x} (a^{7} + (cos (x))^{7})

x = y^{2} + 4 y - 22, x = 3 y + 8

x - tan (x)

x (x + y) \frac{d y}{d x} = y (x - y)

What is (\partial)/(\partial z)((4x-4z)/(y+4z)) ?
The answer to (\partial)/(\partial z)((4x-4z)/(y+4z)) is (-16x-4y)/((y+4z)^2)