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

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

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

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

- 3 x^{2} y^{2} + 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 - 3 x^{2} y^{2} + 2 y

Simplify

= - 3 x^{2} y^{2} + 2 y

\frac{d}{d x} (9^{2})

y = (1 + 2 x)^{2}, (2, 25)

y^{^{''}} + y^{^{'}} - 2 y = 1 - 2 x^{2}

t \to 1 lim (sin (π t))

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