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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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x \to 1 lim (\frac{1}{x ^{2} - x})

18 cos (2 x) - sin (x)

\int_{1}^{2} 7 r^{4} ln (r) d r

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