What is (\partial)/(\partial x)(x^4+3xy^2-3y^2) ?

The answer to (\partial)/(\partial x)(x^4+3xy^2-3y^2) is 4x^3+3y^2

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(\partial)/(\partial x)(x^4+3xy^2-3y^2)

\frac{\partial}{\partial x} (x^{4} + 3 x y^{2} - 3 y^{2})

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

Solution steps

\frac{\partial}{\partial x} (x^{4} + 3 x y^{2} - 3 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4}) + \frac{\partial}{\partial x} (3 x y^{2}) - \frac{\partial}{\partial x} (3 y^{2})

\frac{\partial}{\partial x} (x^{4}) = 4 x^{3}

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 4 x^{3} + 3 y^{2} - 0

Simplify

= 4 x^{3} + 3 y^{2}

What is (\partial)/(\partial x)(x^4+3xy^2-3y^2) ?
The answer to (\partial)/(\partial x)(x^4+3xy^2-3y^2) is 4x^3+3y^2