What is (\partial)/(\partial x)(sqrt(33-3x^2-3y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(33-3x^2-3y^2)) is -(3x)/(sqrt(33-3x^2-3y^2))

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(\partial)/(\partial x)(sqrt(33-3x^2-3y^2))

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

- \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

Solution steps

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} \frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

= \frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} \frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2}) = - 6 x

= \frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} (- 6 x)

Simplify

\frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} (- 6 x) : - \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

= - \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

f (x) = \frac{x}{x ^{2} - 1}, at x = 2

(5 x^{6} + 2 x^{3})^{4}

f (x) = (5 + x) (12 - x) - 2.2 x

\frac{d}{d x} (sec^{2} (π x))

y = x^{2} + 1

What is (\partial)/(\partial x)(sqrt(33-3x^2-3y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(33-3x^2-3y^2)) is -(3x)/(sqrt(33-3x^2-3y^2))