What is (\partial)/(\partial x)(4x^2-2xy+y^2z-y^3) ?

The answer to (\partial)/(\partial x)(4x^2-2xy+y^2z-y^3) is 8x-2y

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(\partial)/(\partial x)(4x^2-2xy+y^2z-y^3)

\frac{\partial}{\partial x} (4 x^{2} - 2 x y + y^{2} z - y^{3})

8 x - 2 y

Solution steps

\frac{\partial}{\partial x} (4 x^{2} - 2 x y + y^{2} z - y^{3})

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 x^{2}) - \frac{\partial}{\partial x} (2 x y) + \frac{\partial}{\partial x} (y^{2} z) - \frac{\partial}{\partial x} (y^{3})

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

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

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

\frac{\partial}{\partial x} (y^{3}) = 0

= 8 x - 2 y + 0 - 0

Simplify

= 8 x - 2 y

\int \frac{3 x ^{2}}{3 + x ^{2}} d x

\frac{3}{2 x + 3 y}

f (x) = 9 x - 2 x^{2}

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

\frac{d y}{d x} = k y (1 - y)

What is (\partial)/(\partial x)(4x^2-2xy+y^2z-y^3) ?
The answer to (\partial)/(\partial x)(4x^2-2xy+y^2z-y^3) is 8x-2y