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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{3 - x ^{2}}{x} d x

8 csc (x) + 2

f (x) = x - 2 sin (x)

n = 0 \sum \infty \frac{1}{2 ^{n} - 1}

\frac{\partial}{\partial x} (z - e^{x y})

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