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 e^{x} cos^{2} (x) d x

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

n = 1 \sum \infty \frac{x ^{n}}{6 ^{n}}

\int tan (x) y d y

ln (14 - 9 x^{2} + 2 x^{4})

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