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

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

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

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

2 x

Solution steps

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

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} (2 z^{2})

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

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

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

= 2 x - 0 - 0

Simplify

= 2 x

\frac{d}{d x} (\frac{12 x ^{5} + 8 x ^{3} - 2 x ^{2}}{4 x ^{2}})

\frac{8}{e ^{x} + e ^{- x}}

e^{- 6 x}, 0

\int x ln (5 x) d x

x \to 2 - lim (6 - 3 x - 3)

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