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

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

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

\int \frac{e ^{4 x}}{1 + e ^{4 x}} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

\frac{\partial}{\partial x} (sin (π (6 x - 6 y)))

\frac{d}{d t} (t + 3)

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