What is (\partial)/(\partial y)(-x^4+4xy-y^4+1) ?

The answer to (\partial)/(\partial y)(-x^4+4xy-y^4+1) is 4x-4y^3

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(\partial)/(\partial y)(-x^4+4xy-y^4+1)

\frac{\partial}{\partial y} (- x^{4} + 4 x y - y^{4} + 1)

4 x - 4 y^{3}

Solution steps

\frac{\partial}{\partial y} (- x^{4} + 4 x y - y^{4} + 1)

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} (4 x y) - \frac{\partial}{\partial y} (y^{4}) + \frac{\partial}{\partial y} (1)

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

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

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

\frac{\partial}{\partial y} (1) = 0

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

Simplify

= 4 x - 4 y^{3}

\frac{\partial}{\partial y} (ln (\frac{y}{x}))

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

x \to 0 + lim (3 + x^{8} ln (x))

\int - \frac{20}{y ^{21}} d y

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

What is (\partial)/(\partial y)(-x^4+4xy-y^4+1) ?
The answer to (\partial)/(\partial y)(-x^4+4xy-y^4+1) is 4x-4y^3