What is (\partial)/(\partial y)(x^4y+xz-yz^2) ?

The answer to (\partial)/(\partial y)(x^4y+xz-yz^2) is x^4-z^2

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(\partial)/(\partial y)(x^4y+xz-yz^2)

\frac{\partial}{\partial y} (x^{4} y + x z - y z^{2})

x^{4} - z^{2}

Solution steps

\frac{\partial}{\partial y} (x^{4} y + x z - y z^{2})

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{4} y) + \frac{\partial}{\partial y} (x z) - \frac{\partial}{\partial y} (y z^{2})

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

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

\frac{\partial}{\partial y} (y z^{2}) = z^{2}

= x^{4} + 0 - z^{2}

Simplify

= x^{4} - z^{2}

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

y = 3 x^{3} - 2 x, (1, 1)

x \to 0 lim (\frac{x + 1}{∣ x ∣})

x \to 0 lim (ln (- x))

\int e^{\frac{y}{x}} d y

What is (\partial)/(\partial y)(x^4y+xz-yz^2) ?
The answer to (\partial)/(\partial y)(x^4y+xz-yz^2) is x^4-z^2