What is the d/(dy)(sqrt(17-x^2-4y^2)) ?

The d/(dy)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))

What is the first d/(dy)(sqrt(17-x^2-4y^2)) ?

The first d/(dy)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))

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

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

Solution steps

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

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

\frac{d}{d 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}}

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

\frac{\partial}{\partial x} (sin (x + x y + z^{2}))

\frac{\partial}{\partial x} (y^{4} cos (5 x))

\int_{5}^{8} \frac{1}{9 + ( x - 5 ) ^{2}} d x

n = 1 \sum \infty 2^{n + 1} - 2^{n}

What is the d/(dy)(sqrt(17-x^2-4y^2)) ?
The d/(dy)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))
What is the first d/(dy)(sqrt(17-x^2-4y^2)) ?
The first d/(dy)(sqrt(17-x^2-4y^2)) is -(4y)/(sqrt(17-x^2-4y^2))