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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= x \frac{\partial}{\partial y} (\frac{1}{4 + y})

Apply exponent rule:

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

= x \frac{\partial}{\partial y} ((4 + y)^{- 1})

Apply the chain rule:

- \frac{1}{( 4 + y ) ^{2}} \frac{\partial}{\partial y} (4 + y)

= - \frac{1}{( 4 + y ) ^{2}} \frac{\partial}{\partial y} (4 + y)

\frac{\partial}{\partial y} (4 + y) = 1

= x (- \frac{1}{( 4 + y ) ^{2}} \cdot 1)

Simplify

x (- \frac{1}{( 4 + y ) ^{2}} \cdot 1) : - \frac{x}{( 4 + y ) ^{2}}

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

81 y^{^{''}} + 180 y^{^{'}} + 125 y = 0, y (3) = 6, y^{^{'}} (3) = 6

\frac{d}{d x} (8 cos^{2} (x))

2 x^{3} - 6 x^{2} - 2 x + 6, - x^{3} + 3 x^{2} + x - 3

y = 5 cos (7 x), y = 5 - 5 cos (7 x)

\int_{0}^{\frac{π}{16}} cos (8 x) d x

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