What is (\partial)/(\partial y)(sqrt(17-x^2-4y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))

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(\partial)/(\partial y)(sqrt(17-x^2-4y^2))

\frac{\partial}{\partial y} (17 - x^{2} - 4 y^{2})

- \frac{4 y}{17 - x ^{2} - 4 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (17 - x^{2} - 4 y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 17 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (17 - x^{2} - 4 y^{2})

= \frac{1}{2 17 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (17 - x^{2} - 4 y^{2})

\frac{\partial}{\partial y} (17 - x^{2} - 4 y^{2}) = - 8 y

= \frac{1}{2 17 - x ^{2} - 4 y ^{2}} (- 8 y)

Simplify

\frac{1}{2 17 - x ^{2} - 4 y ^{2}} (- 8 y) : - \frac{4 y}{17 - x ^{2} - 4 y ^{2}}

= - \frac{4 y}{17 - x ^{2} - 4 y ^{2}}

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

\int \frac{1}{2} (1 - cos (x))^{2} d x

\int_{0}^{1} x 4 + 5 x d x

\frac{d}{d x} (cos (x) ln (x^{2} + 1) - ln (x^{4} + 1))

y^{^{'}} = - y^{3}, y (0) = a

What is (\partial)/(\partial y)(sqrt(17-x^2-4y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))