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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

ln (cos (x))

\frac{d y}{d x} = \frac{x}{36 y}

\frac{8}{s ^{2} + 64}

\int (4 x^{3} + 6 x^{2} - 1) d x

\int csc (\frac{x}{2}) d x

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