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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{\partial}{\partial y} ((2 x + 3 y)^{2}) = 6 (2 x + 3 y)

= - 2 \cdot \frac{1 \cdot ( 2 x + 3 y ) ^{2} - 6 ( 2 x + 3 y ) y}{( ( 2 x + 3 y ) ^{2} ) ^{2}}

Simplify

- 2 \cdot \frac{1 \cdot ( 2 x + 3 y ) ^{2} - 6 ( 2 x + 3 y ) y}{( ( 2 x + 3 y ) ^{2} ) ^{2}} : - \frac{2 ( 2 x - 3 y )}{( 2 x + 3 y ) ^{3}}

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

\frac{d}{d x} (- \frac{3}{( 1 + 3 x ) ^{2}})

(x^{2} - 2 x + 5) 25 - x^{2}

- 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\int 3 x + \frac{1}{x} d x

\frac{s ^{2} - 1}{s ( s + 2 ) ( s - 3 )}

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