What is the derivative of (-1/(x^2+y^2)) ?

The derivative of (-1/(x^2+y^2)) is (2x+2y\frac{dy)/(dx)}{(x^2+y^2)^2}

What is the first derivative of (-1/(x^2+y^2)) ?

The first derivative of (-1/(x^2+y^2)) is (2x+2y\frac{dy)/(dx)}{(x^2+y^2)^2}

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the derivative of (-1/(x^2+y^2)) ?
The derivative of (-1/(x^2+y^2)) is (2x+2y\frac{dy)/(dx)}{(x^2+y^2)^2}
What is the first derivative of (-1/(x^2+y^2)) ?
The first derivative of (-1/(x^2+y^2)) is (2x+2y\frac{dy)/(dx)}{(x^2+y^2)^2}