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

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

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

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

4 x^{3} + 4 y

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 x^{3} + 4 y

\frac{d}{d x} (y - x)

x \to 1 lim (\frac{x ^{2}}{x ^{3} - x})

\frac{d}{d x} (cos (x) tan (x))

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

\frac{1}{6} y^{3} + \frac{1}{2 y}

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