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

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

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

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

2 x

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (4) = 0

= 2 x - 0 + 0 - 0 - 0

Simplify

= 2 x

f (x) = x^{2} - 8, (- 5, 17)

\frac{d y}{d x} = \frac{x ^{2} + 2}{y}

x \to - 3 lim (- 6)

(- 1.2) (- 5.2)

\int - 17 sin^{4} (x) cos (x) d x

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