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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= - 2 \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

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

Simplify

- 2 \cdot \frac{1}{2 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} (x^{- \frac{4}{3}})

y^{^{'}} + k y = e^{- 7 k x}

\frac{1}{64 + 5 t ^{2}}

\int \frac{1}{x ^{2} + 9} d x

f (x) = 2 x^{3} - x^{2} + 3 x, at x = 1

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