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

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

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

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

4 x^{3} - 6 x y^{2}

Solution steps

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

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

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

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

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

= 4 x^{3} - 6 y^{2} x + 0

Simplify

= 4 x^{3} - 6 x y^{2}

x \to - 2 lim (4^{2} - 3)

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

f (x) = 4 x^{2} - 1

y = (1 + 5 x)^{12}, (0, 1)

x y^{^{'}} + 2 y = 2 e^{x}

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