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)/((x^2+y^2)^2)

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (x^{2} + y^{2})^{- 1}

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

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

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

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

Simplify

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

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

f (x) = 4 x cos (x)

x - 9

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{3 t}

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