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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

x^{2} 25 - x^{2}

sin (2 t + \frac{π}{4})

\int (5 x + 2) d x

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

\int \frac{- x}{1 - x ^{2}} d x

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