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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

t^{2} y^{^{''}} - 2 y = 3 t^{2} - 1

x \to 0 lim (\frac{( 1 )}{x})

y^{^{'}} = (1 - y) (2 - y)

(ln (e))^{^{'}}

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

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