What is (\partial)/(\partial y)(sqrt(38-x^2-4y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(38-x^2-4y^2)) is -(4y)/(sqrt(38-x^2-4y^2))

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(\partial)/(\partial y)(sqrt(38-x^2-4y^2))

\frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})

- \frac{4 y}{38 - x ^{2} - 4 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 38 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})

= \frac{1}{2 38 - x ^{2} - 4 y ^{2}} \frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})

\frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2}) = - 8 y

= \frac{1}{2 38 - x ^{2} - 4 y ^{2}} (- 8 y)

Simplify

\frac{1}{2 38 - x ^{2} - 4 y ^{2}} (- 8 y) : - \frac{4 y}{38 - x ^{2} - 4 y ^{2}}

= - \frac{4 y}{38 - x ^{2} - 4 y ^{2}}

x^{4} + y^{3} x^{5} + 4 \frac{d y}{d x} = 0, y (1) = 0

\int \frac{t ^{5}}{t ^{2} + 8} d t

\frac{d}{d x} (e^{k t a n (3 x)})

\int - 64 x^{3} sin (4 x^{4} + 3) d x

h \to 0 lim (\frac{( 1 + h ) ^{7} - 1}{h})

What is (\partial)/(\partial y)(sqrt(38-x^2-4y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(38-x^2-4y^2)) is -(4y)/(sqrt(38-x^2-4y^2))