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

e^{- t} (\frac{e ^{t} + e ^{- t}}{2})

f (x) = \frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

n \to \infty lim (4 n)

\frac{\partial}{\partial x} (2 x^{\frac{1}{2}} y^{\frac{1}{3}})

y^{^{'}} = \frac{y}{2}

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