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} )})

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}}

\frac{\partial}{\partial x} (e^{7 x y})

\int sin^{3} (x) - sin^{5} (x) d x

f (x) = \frac{x}{x ^{2} - 4}

9 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

y^{^{'}} = y (x y^{5} + 1)

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}