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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x^{3} - 2 x - 5)

x \to 0 lim (x^{4} ln (x))

(- 2.4) (2.4)

\frac{\partial}{\partial x} (2^{x^{2}})

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