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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

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

\int \frac{cos ( x )}{sin ( x ) + 1} d x

\frac{d}{d x} (\frac{sin ( x )}{2})

y = \frac{4 x}{x ^{2} + 1}

f (x) = x^{2.4} + e^{2.4}

\frac{\partial}{\partial x} (5 x^{4} y - 3 y sin (x))

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