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}

\int_{0}^{4} (1 + 3 x - x^{2}) d x

\int_{- \infty}^{0} \frac{1}{3 - 10 x} d x

- 4 x^{5} + 6 x^{7}

x \to + 0 lim (\frac{e ^{x} - 1}{5 x})

\int \frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} d x

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