What is (\partial)/(\partial y)(x^2+2y^2-1) ?

The answer to (\partial)/(\partial y)(x^2+2y^2-1) is 4y

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(\partial)/(\partial y)(x^2+2y^2-1)

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

4 y

Solution steps

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

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} (2 y^{2}) - \frac{\partial}{\partial y} (1)

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

\frac{\partial}{\partial y} (2 y^{2}) = 4 y

\frac{\partial}{\partial y} (1) = 0

= 0 + 4 y - 0

Simplify

= 4 y

\int \frac{9 x ^{8} + 8}{x ^{9} + 8 x} d x

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

8 y^{^{''}} + 32 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

\frac{x ^{2}}{2 + x}

\frac{d x}{d y} = (4 x + 2 y + 1)

What is (\partial)/(\partial y)(x^2+2y^2-1) ?
The answer to (\partial)/(\partial y)(x^2+2y^2-1) is 4y