What is (\partial)/(\partial y)(x/(5+y)) ?

The answer to (\partial)/(\partial y)(x/(5+y)) is -x/((5+y)^2)

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(\partial)/(\partial y)(x/(5+y))

\frac{\partial}{\partial y} (\frac{x}{5 + y})

- \frac{x}{( 5 + y ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{x}{5 + y})

Treat

x

as a constant

Take the constant out:

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

= x \frac{\partial}{\partial y} (\frac{1}{5 + y})

Apply exponent rule:

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

= x \frac{\partial}{\partial y} ((5 + y)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 + y ) ^{2}} \frac{\partial}{\partial y} (5 + y)

= - \frac{1}{( 5 + y ) ^{2}} \frac{\partial}{\partial y} (5 + y)

\frac{\partial}{\partial y} (5 + y) = 1

= x (- \frac{1}{( 5 + y ) ^{2}} \cdot 1)

Simplify

x (- \frac{1}{( 5 + y ) ^{2}} \cdot 1) : - \frac{x}{( 5 + y ) ^{2}}

= - \frac{x}{( 5 + y ) ^{2}}

x \to - 1 lim (x - 1)

\frac{d ^{2}}{d t ^{2}} x + 5 \frac{d}{d t} x + 2 x = 0

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

x \to 0 lim (\frac{7 ^{x} - 2 ^{x}}{x})

\frac{d y}{d x} + 2 x y + x y^{4} = 0

What is (\partial)/(\partial y)(x/(5+y)) ?
The answer to (\partial)/(\partial y)(x/(5+y)) is -x/((5+y)^2)