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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{\partial}{\partial x} (6 e^{x^{3} y^{2}})

5 x^{2} + 5 x + 2

\int x^{\frac{3}{2}} (x^{2} + 3) (x + 2) d x

y^{^{'''}} + y^{^{''}} - 2 y = 0

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