What is (\partial)/(\partial y)(sqrt(4-x^2-8y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(4-x^2-8y^2)) is -(8y)/(sqrt(4-x^2-8y^2))

Popular >

(\partial)/(\partial y)(sqrt(4-x^2-8y^2))

\frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2})

- \frac{8 y}{4 - x ^{2} - 8 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 4 - x ^{2} - 8 y ^{2}} \frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2})

= \frac{1}{2 4 - x ^{2} - 8 y ^{2}} \frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2})

\frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2}) = - 16 y

= \frac{1}{2 4 - x ^{2} - 8 y ^{2}} (- 16 y)

Simplify

\frac{1}{2 4 - x ^{2} - 8 y ^{2}} (- 16 y) : - \frac{8 y}{4 - x ^{2} - 8 y ^{2}}

= - \frac{8 y}{4 - x ^{2} - 8 y ^{2}}

y = \frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (3 x, - 2 y, x - y)

\frac{d}{d x} ((e^{- x^{2}}) (- 2 x))

y^{^{''}} + y = - sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

\int \frac{ln ( x ^{20} )}{x} d x

What is (\partial)/(\partial y)(sqrt(4-x^2-8y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(4-x^2-8y^2)) is -(8y)/(sqrt(4-x^2-8y^2))