What is (\partial)/(\partial y)(xyz+xy^4+yz^4+zx^4) ?

The answer to (\partial)/(\partial y)(xyz+xy^4+yz^4+zx^4) is xz+4xy^3+z^4

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(\partial)/(\partial y)(xyz+xy^4+yz^4+zx^4)

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

x z + 4 x y^{3} + z^{4}

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x yz) = x z

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

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

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

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

Simplify

= x z + 4 x y^{3} + z^{4}

\frac{3}{s ^{2} + 2 s + 1}

y = - x^{3}, (1, - 1)

\int \frac{arctan ( x )}{x} d x

\int \frac{6 ln ( x )}{x} d x

\frac{d}{d x} (\frac{1}{9 + x ^{2}})

What is (\partial)/(\partial y)(xyz+xy^4+yz^4+zx^4) ?
The answer to (\partial)/(\partial y)(xyz+xy^4+yz^4+zx^4) is xz+4xy^3+z^4