What is (\partial)/(\partial x)((1x-1y)/(1x+1y)) ?

The answer to (\partial)/(\partial x)((1x-1y)/(1x+1y)) is (2y)/((x+y)^2)

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(\partial)/(\partial x)((1x-1y)/(1x+1y))

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

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

Solution steps

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

\frac{\partial}{\partial x} (1 \cdot x - 1 \cdot y) = 1

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

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

Simplify

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

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

n = 0 \sum \infty (- 1)^{n} \frac{( ln ( 2 ) ) ^{n}}{n !}

n = 2 \sum \infty \frac{1}{3 ^{n} - 2}

\frac{d}{d x} (18 x a - 8 + 32 x a - b + 15 + 18 x c - b + 16)

\int cos (x) e^{s i n (x)} d x

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

What is (\partial)/(\partial x)((1x-1y)/(1x+1y)) ?
The answer to (\partial)/(\partial x)((1x-1y)/(1x+1y)) is (2y)/((x+y)^2)