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

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

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

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

4 x - 4 y^{3}

Solution steps

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

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} (1)

\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}

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

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

Simplify

= 4 x - 4 y^{3}

n = 2 \sum \infty \frac{( 3 x - 2 ) ^{n}}{n}

g (x) = 5 x^{2} tan (1 - x)

\int \frac{x}{x ^{2} - 15} d x

y = sin (2 t)

\int 4 cos^{4} (5 x) d x

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