What is (\partial)/(\partial x)(x^2+y^2+z^2-9) ?

The answer to (\partial)/(\partial x)(x^2+y^2+z^2-9) is 2x

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(\partial)/(\partial x)(x^2+y^2+z^2-9)

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 9)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 9)

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} (9)

\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} (9) = 0

= 2 x + 0 + 0 - 0

Simplify

= 2 x

\int \frac{e ^{x}}{1 - 64 e ^{2 x}} d x

\frac{\partial}{\partial x} (\frac{1}{ln ( x - y )})

n = 1 \sum \infty \frac{1}{n ^{0.3}}

\int - e^{- y} d y

f (x) = x^{2} \cdot sin (x)

What is (\partial)/(\partial x)(x^2+y^2+z^2-9) ?
The answer to (\partial)/(\partial x)(x^2+y^2+z^2-9) is 2x