What is (\partial)/(\partial y)(sqrt(26-x^2-4y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(26-x^2-4y^2)) is -(4y)/(sqrt(26-x^2-4y^2))

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(\partial)/(\partial y)(sqrt(26-x^2-4y^2))

\frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

- \frac{4 y}{26 - x ^{2} - 4 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 26 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

= \frac{1}{2 26 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

\frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2}) = - 8 y

= \frac{1}{2 26 - x ^{2} - 4 y ^{2}} (- 8 y)

Simplify

\frac{1}{2 26 - x ^{2} - 4 y ^{2}} (- 8 y) : - \frac{4 y}{26 - x ^{2} - 4 y ^{2}}

= - \frac{4 y}{26 - x ^{2} - 4 y ^{2}}

\int \frac{6 x}{7 - 3 x ^{2}} d x

\int_{0}^{2} (4 - x^{2})^{2} d x

x \to 1 - lim (4)

y = \frac{1}{x - 5}, at x = 10

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

What is (\partial)/(\partial y)(sqrt(26-x^2-4y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(26-x^2-4y^2)) is -(4y)/(sqrt(26-x^2-4y^2))