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

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

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

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

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 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} (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} (4) = 0

= 2 x + 0 + 0 - 0

Simplify

= 2 x

y (θ) = \frac{π}{3} sin (θ)

\frac{\partial}{\partial x} (9 x cos (3 x y))

\frac{d}{d x} ((x + a)^{2})

4 t

sin^{2} (5 t)

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