What is (\partial)/(\partial y)(x^2+y^2-9) ?

The answer to (\partial)/(\partial y)(x^2+y^2-9) is 2y

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(\partial)/(\partial y)(x^2+y^2-9)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 9)

2 y

Solution steps

\frac{\partial}{\partial y} (x^{2} + y^{2} - 9)

Treat

x

as a constant

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} (9)

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

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

\frac{\partial}{\partial y} (9) = 0

= 0 + 2 y - 0

Simplify

= 2 y

f (x) = 6 x^{2} + x - 3

\frac{t}{2}

y = x, x = 9, y = 0

x y^{^{'}} + 2 y = 8 x^{2}

3 y^{^{''}} + 5 y^{^{'}} + 2 y = 0

What is (\partial)/(\partial y)(x^2+y^2-9) ?
The answer to (\partial)/(\partial y)(x^2+y^2-9) is 2y