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

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

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(\partial)/(\partial x)((x-y)/(x^2+y+1))

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

\int_{0}^{3} (2 - x) e^{x} d x

\frac{8}{( s + 1 ) ^{2} ( s + 2 )}

\int \frac{1}{( x - 4 ) ( x - 1 )} d x

\int_{2}^{3} \frac{7}{3 - x} d x

\frac{d}{d x} (\frac{( x ^{2} - 3 x - 4 )}{x})

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