What is (\partial)/(\partial y)(x^2-3y^2+7) ?

The answer to (\partial)/(\partial y)(x^2-3y^2+7) is -6y

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(\partial)/(\partial y)(x^2-3y^2+7)

\frac{\partial}{\partial y} (x^{2} - 3 y^{2} + 7)

- 6 y

Solution steps

\frac{\partial}{\partial y} (x^{2} - 3 y^{2} + 7)

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} (3 y^{2}) + \frac{\partial}{\partial y} (7)

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

\frac{\partial}{\partial y} (3 y^{2}) = 6 y

\frac{\partial}{\partial y} (7) = 0

= 0 - 6 y + 0

Simplify

= - 6 y

\int \frac{( x + 1 )}{( x - 1 )} d x

x \to 1 + lim (2)

f (x) = x, (16, 4)

x \to 5 lim (2 x^{2} - 5 x + 3)

y = x + 8

What is (\partial)/(\partial y)(x^2-3y^2+7) ?
The answer to (\partial)/(\partial y)(x^2-3y^2+7) is -6y