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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

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

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

\frac{d}{d x} (2 + ln (x))

\int \frac{361 - x ^{2}}{x} d x

\frac{d}{d x} (a cos (2 x))

\int 8 cos (θ) d θ

y = cos (x), y = sin (2 x), x = 0, x = \frac{π}{2}

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