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

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

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

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

2 x y - 2 y

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (2 x y) = 2 y

\frac{\partial}{\partial x} (y z^{2}) = 0

= 2 y x - 2 y + 0

Simplify

= 2 x y - 2 y

(- 2, 3), (4, - 5)

n = 0 \sum \infty \frac{e ^{n}}{n !}

\frac{\partial}{\partial x} (10 x e^{- x^{2} - y^{2}})

y = x^{3} - 16 x, y = 4 - x, 0, 2

θ cos (θ) sin (θ)

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