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

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

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

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

4 x - 4 y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= 4 x - 4 y^{3}

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

\frac{\partial}{\partial x} (\frac{9 y}{x ^{4}})

x \to 2 lim (- 3)

\int sec^{2} (x) (1 - cot^{2} (x)) d x

x \to 0 lim ((1 + 5 x)^{\frac{1}{x}})

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