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

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

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

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

2 x

Solution steps

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

Treat

y, z, t

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} (t^{2})

\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} (t^{2}) = 0

= 2 x - 0 + 0 - 0

Simplify

= 2 x

- 2 x^{2} + 8, 3 x + 1

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

y^{^{'}} = (4 y + 21)^{(\frac{1}{2})}

\frac{d y}{d x} = - y

\frac{\partial}{\partial x} (2 y - 2 cos (x))

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