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

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

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

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

4 y - 4 x^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= 4 y - 4 x^{3}

y = x^{3}, y = 2 x - x^{2}

x \to 2 lim (\frac{( x ^{2} + 7 x - 44 )}{x ^{2} - 6 x + 8})

\int_{- \infty}^{- 1} e^{- 34 x} d x

\int \frac{sin ( x )}{cos ^{9} ( x )} d x

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