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

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

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

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

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

Solution steps

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

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

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

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

Expand

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

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

y = 2 - x^{2}, y = 6 - 5 x

f (x) = 2 x - 3 x^{\frac{2}{3}}

y (X) = X^{2} + X^{- 1}

\frac{\partial}{\partial x} (x cos (x + 2 y))

\frac{\partial}{\partial x} (ln (\frac{y}{cos ( x )}))

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