What is (\partial)/(\partial x)(((2x-4y))/(2x+4y)) ?

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

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(\partial)/(\partial x)(((2x-4y))/(2x+4y))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int sin^{- 2} (x) d x

\frac{d}{d x} (20 - x^{2})

x \to 6 lim (f (x))

\int \frac{t}{1 - t ^{3}} d t

\frac{d y}{d x} = x + 4 y

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