What is (\partial)/(\partial x)(x^2+y^2+z^2-1) ?

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

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(\partial)/(\partial x)(x^2+y^2+z^2-1)

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 1)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 1)

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} (1)

\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} (1) = 0

= 2 x + 0 + 0 - 0

Simplify

= 2 x

arccos (\frac{cos ( x )}{1 + cos ( x )})

f (x) = sin (8 x)

y = x^{9 c o s (x)}

\int 2 - ∣ x ∣ d x

f (x) = \frac{1}{x}, (2, \frac{1}{2})

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