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

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

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

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

- 2 x

Solution steps

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

Simplify

- (x^{2} + y^{2}) : - x^{2} - y^{2}

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial x} (y^{2}) = 0

= - 2 x - 0

Simplify

= - 2 x

x \to 1 - lim (2 x + 2)

- 3 x^{2} - 8 x

y^{^{''}} + 8 y^{^{'}} + 16 y = 0

\frac{d y}{d x} = x^{5} y

\frac{d}{d t} (\frac{t ^{2} + 2}{t ^{4} - 3 t ^{2} + 1})

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