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

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

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

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

2 x

Solution steps

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

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} (2 yz)

\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} (2 yz) = 0

= 2 x + 0 + 0 - 0

Simplify

= 2 x

\frac{d}{d x} (d y (d, x))

\frac{7}{x}, 4

\int_{1}^{4} x ln (\frac{x}{2}) d x

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

\int_{- 5}^{1} (2 x + 6) d x

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