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

The answer to (\partial)/(\partial x)(x-2y^2-2z^2) is 1

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

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

1

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

\frac{\partial}{\partial x} (x) = 1

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

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

= 1 - 0 - 0

Simplify

= 1

\int_{1}^{9} \frac{x - 1}{x} d x

x \to 2 lim (3 x^{2} - 4 x + 2)

x^{3} (x - 3)^{2}

\frac{d}{d x} (\frac{x}{x ^{3} - 1})

\int (u + 4) (2 u + 1) d u

What is (\partial)/(\partial x)(x-2y^2-2z^2) ?
The answer to (\partial)/(\partial x)(x-2y^2-2z^2) is 1