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

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

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

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

2 z x

Solution steps

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

Treat

z, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

= z (2 x + 0 - 0)

Simplify

= 2 z x

4^{3}

x^{2} - 2 (a + \frac{1}{a}) x + 4

\frac{9 ( x - 2 ) ^{2}}{4}

(sin (2 θ) + cos (2 θ)) (sin (θ) + cos (θ))

\frac{34 ( s + 4 )}{( s ^{2} + 2 s + 2 ) ( s ^{2} + 16 s + 68 )}

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