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

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

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

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

2 z

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 0 + 2 z

Simplify

= 2 z

x \to 0 + lim (x^{s i n (2 x)})

y = 6 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (sin (2 x) cos (3 x))

\frac{\partial}{\partial x} (z^{x y})

y^{^{''}} + 9 y = 4 csc^{2} (3 x)

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