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

The answer to (\partial)/(\partial x)(A(B-x^2-y^2-z^2)) is -2Ax

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(\partial)/(\partial x)(A(B-x^2-y^2-z^2))

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

- 2 A x

Solution steps

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

Treat

A, B, y, z

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (B) = 0

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

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

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

= A (0 - 2 x - 0 - 0)

Simplify

= - 2 A x

f (x) = (5 x^{6} + 4 x^{3})^{4}

\frac{d}{d x} (x + 2 x^{2} sin (\frac{1}{x}))

y = - 2 x^{2} + 4 x - 4

x \to - \infty lim (ln (x))

f (x) = \frac{x ^{2}}{5 + 8 x}

What is (\partial)/(\partial x)(A(B-x^2-y^2-z^2)) ?
The answer to (\partial)/(\partial x)(A(B-x^2-y^2-z^2)) is -2Ax