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

The answer to (\partial)/(\partial x)(x^4+y^4+3x^2y^3) is 4x^3+6xy^3

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

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

4 x^{3} + 6 x y^{3}

Solution steps

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

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

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

\frac{\partial}{\partial x} (y^{4}) = 0

\frac{\partial}{\partial x} (3 x^{2} y^{3}) = 6 y^{3} x

= 4 x^{3} + 0 + 6 y^{3} x

Simplify

= 4 x^{3} + 6 x y^{3}

13 + 2 x - x^{2}, 3 x + 1

f (x) = (x + 1)^{2 x}

\frac{x}{a}

\frac{x}{y} = sin (\frac{π x y}{2}), (1, 1)

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

What is (\partial)/(\partial x)(x^4+y^4+3x^2y^3) ?
The answer to (\partial)/(\partial x)(x^4+y^4+3x^2y^3) is 4x^3+6xy^3