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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (- e^{- 5 x})

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

\int_{0}^{1} \frac{1}{2 x - 1} d x

\frac{\partial}{\partial x} (sec (θ))

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