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

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

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

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

2 y

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 2 y + 0

Simplify

= 2 y

\int 5 cos^{3} (2 x) d x

\int \frac{ln ( x ^{3} )}{- 804 x ^{2}} d x

(0.3, - 1), (- 0.1, - 0.8)

\frac{\partial}{\partial x} (8 e^{x})

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

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