What is (\partial)/(\partial x)(20-(3x^2+y^2)) ?

The answer to (\partial)/(\partial x)(20-(3x^2+y^2)) is -6x

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(\partial)/(\partial x)(20-(3x^2+y^2))

\frac{\partial}{\partial x} (20 - (3 x^{2} + y^{2}))

- 6 x

Solution steps

\frac{\partial}{\partial x} (20 - (3 x^{2} + y^{2}))

Simplify

20 - (3 x^{2} + y^{2}) : 20 - 3 x^{2} - y^{2}

= \frac{\partial}{\partial x} (20 - 3 x^{2} - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (20) - \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (20) = 0

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

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

= 0 - 6 x - 0

Simplify

= - 6 x

\frac{d}{d x} (\frac{5 x}{2})

\frac{\partial}{\partial y} (- e^{x} ln (y))

y = sin (sin (x)), (π, 0)

\frac{d}{d x} (\frac{8 x ^{2}}{2 x - 10})

\frac{d}{d x} (\frac{x}{t})

What is (\partial)/(\partial x)(20-(3x^2+y^2)) ?
The answer to (\partial)/(\partial x)(20-(3x^2+y^2)) is -6x