What is (\partial)/(\partial x)(1-9x^2-6y^2) ?

The answer to (\partial)/(\partial x)(1-9x^2-6y^2) is -18x

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(\partial)/(\partial x)(1-9x^2-6y^2)

\frac{\partial}{\partial x} (1 - 9 x^{2} - 6 y^{2})

- 18 x

Solution steps

\frac{\partial}{\partial x} (1 - 9 x^{2} - 6 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (1) - \frac{\partial}{\partial x} (9 x^{2}) - \frac{\partial}{\partial x} (6 y^{2})

\frac{\partial}{\partial x} (1) = 0

\frac{\partial}{\partial x} (9 x^{2}) = 18 x

\frac{\partial}{\partial x} (6 y^{2}) = 0

= 0 - 18 x - 0

Simplify

= - 18 x

f (x) = (x + 2) \cdot - x

\frac{d y}{d x} = \frac{2 y}{x} - x^{2} y^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{3 + e ^{x}}

d y - (y - 6)^{2} d x = 0

What is (\partial)/(\partial x)(1-9x^2-6y^2) ?
The answer to (\partial)/(\partial x)(1-9x^2-6y^2) is -18x