What is (\partial)/(\partial x)(7-3x^2-3y^2) ?

The answer to (\partial)/(\partial x)(7-3x^2-3y^2) is -6x

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(\partial)/(\partial x)(7-3x^2-3y^2)

\frac{\partial}{\partial x} (7 - 3 x^{2} - 3 y^{2})

- 6 x

Solution steps

\frac{\partial}{\partial x} (7 - 3 x^{2} - 3 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (7) - \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (3 y^{2})

\frac{\partial}{\partial x} (7) = 0

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 0 - 6 x - 0

Simplify

= - 6 x

(2.4) (6.12)

x \to 0 + lim ((x + 1)^{in x})

6 x + 5 cos (x)

x \to \infty lim (x^{2} + x)

n = 0 \sum \infty \frac{5 ^{2 n}}{n !}

What is (\partial)/(\partial x)(7-3x^2-3y^2) ?
The answer to (\partial)/(\partial x)(7-3x^2-3y^2) is -6x