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

The answer to (\partial)/(\partial x)(x^4+y^4-4x^2y^2) is 4x^3-8xy^2

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

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

4 x^{3} - 8 x y^{2}

Solution steps

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

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

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

\frac{\partial}{\partial x} (4 x^{2} y^{2}) = 8 y^{2} x

= 4 x^{3} + 0 - 8 y^{2} x

Simplify

= 4 x^{3} - 8 x y^{2}

\frac{\partial}{\partial y} (\frac{y}{x ^{2} y ^{2} + 1})

x^{2} + x^{3}

\int e^{t - s} d s

\frac{d}{d x} (2 π^{2})

What is (\partial)/(\partial x)(x^4+y^4-4x^2y^2) ?
The answer to (\partial)/(\partial x)(x^4+y^4-4x^2y^2) is 4x^3-8xy^2