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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} = (x - 4) e^{- 2 y}

\frac{d}{d x} (\frac{- 2 x ^{2} + 8}{8 - x ^{2}})

\int_{0}^{\frac{π}{6}} sec (x) d x

x \to - 1 lim (\frac{1}{- x - 1})

\int (2 x^{2} - 4 x)^{9} (x - 1) d x

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