What is (\partial)/(\partial y)(5x^2-2xy+3y^3) ?

The answer to (\partial)/(\partial y)(5x^2-2xy+3y^3) is 9y^2-2x

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(\partial)/(\partial y)(5x^2-2xy+3y^3)

\frac{\partial}{\partial y} (5 x^{2} - 2 x y + 3 y^{3})

9 y^{2} - 2 x

Solution steps

\frac{\partial}{\partial y} (5 x^{2} - 2 x y + 3 y^{3})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (5 x^{2}) - \frac{\partial}{\partial y} (2 x y) + \frac{\partial}{\partial y} (3 y^{3})

\frac{\partial}{\partial y} (5 x^{2}) = 0

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

\frac{\partial}{\partial y} (3 y^{3}) = 9 y^{2}

= 0 - 2 x + 9 y^{2}

Simplify

= 9 y^{2} - 2 x

(5.8) (3.9)

y^{^{''}} - y = \frac{1}{e ^{x} + e ^{- x}}

\frac{\partial}{\partial x} (cos (x y))

\frac{d}{d x} (\frac{tan ( x )}{1 + cos ( x )})

\int_{2}^{3} \frac{36}{3 - x} d x

What is (\partial)/(\partial y)(5x^2-2xy+3y^3) ?
The answer to (\partial)/(\partial y)(5x^2-2xy+3y^3) is 9y^2-2x