What is (\partial)/(\partial x)(sqrt(16-x^2-4y)) ?

The answer to (\partial)/(\partial x)(sqrt(16-x^2-4y)) is -x/(sqrt(16-x^2-4y))

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(\partial)/(\partial x)(sqrt(16-x^2-4y))

\frac{\partial}{\partial x} (16 - x^{2} - 4 y)

- \frac{x}{16 - x ^{2} - 4 y}

Solution steps

\frac{\partial}{\partial x} (16 - x^{2} - 4 y)

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 16 - x ^{2} - 4 y} \frac{\partial}{\partial x} (16 - x^{2} - 4 y)

= \frac{1}{2 16 - x ^{2} - 4 y} \frac{\partial}{\partial x} (16 - x^{2} - 4 y)

\frac{\partial}{\partial x} (16 - x^{2} - 4 y) = - 2 x

= \frac{1}{2 16 - x ^{2} - 4 y} (- 2 x)

Simplify

\frac{1}{2 16 - x ^{2} - 4 y} (- 2 x) : - \frac{x}{16 - x ^{2} - 4 y}

= - \frac{x}{16 - x ^{2} - 4 y}

x \to \infty lim (\frac{2}{x !})

\frac{( csc ^{2} ( x ) )}{4}, 4 sin^{2} (x)

\frac{d ^{4}}{d x ^{4}} (cos (x))

\int v d v

x \to 10 lim (5)

What is (\partial)/(\partial x)(sqrt(16-x^2-4y)) ?
The answer to (\partial)/(\partial x)(sqrt(16-x^2-4y)) is -x/(sqrt(16-x^2-4y))