What is (\partial)/(\partial y)((2x-2z)/(3y+2z)) ?

The answer to (\partial)/(\partial y)((2x-2z)/(3y+2z)) is -(3(2x-2z))/((3y+2z)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)((2x-2z)/(3y+2z))

\frac{\partial}{\partial y} (\frac{2 x - 2 z}{3 y + 2 z})

- \frac{3 ( 2 x - 2 z )}{( 3 y + 2 z ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{2 x - 2 z}{3 y + 2 z})

Treat

x, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= (2 x - 2 z) \frac{\partial}{\partial y} (\frac{1}{3 y + 2 z})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= (2 x - 2 z) \frac{\partial}{\partial y} ((3 y + 2 z)^{- 1})

Apply the chain rule:

- \frac{1}{( 3 y + 2 z ) ^{2}} \frac{\partial}{\partial y} (3 y + 2 z)

= - \frac{1}{( 3 y + 2 z ) ^{2}} \frac{\partial}{\partial y} (3 y + 2 z)

\frac{\partial}{\partial y} (3 y + 2 z) = 3

= (2 x - 2 z) (- \frac{1}{( 3 y + 2 z ) ^{2}} \cdot 3)

Simplify

(2 x - 2 z) (- \frac{1}{( 3 y + 2 z ) ^{2}} \cdot 3) : - \frac{3 ( 2 x - 2 z )}{( 3 y + 2 z ) ^{2}}

= - \frac{3 ( 2 x - 2 z )}{( 3 y + 2 z ) ^{2}}

y = 2 x^{2} - 4 x + 2

\int \frac{( ln ( x ) ) ^{25}}{x} d x

\int_{1}^{2} w^{5} e^{w^{3}} d w

\frac{d y}{d x} - e^{2 x} = 3, y (0) = 4

x \to - 2 lim (\frac{x ^{2}}{x ^{2} + 2})

What is (\partial)/(\partial y)((2x-2z)/(3y+2z)) ?
The answer to (\partial)/(\partial y)((2x-2z)/(3y+2z)) is -(3(2x-2z))/((3y+2z)^2)