What is (\partial)/(\partial x)(sqrt(15-x^2-2y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(15-x^2-2y^2)) is -x/(sqrt(15-x^2-2y^2))

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(\partial)/(\partial x)(sqrt(15-x^2-2y^2))

\frac{\partial}{\partial x} (15 - x^{2} - 2 y^{2})

- \frac{x}{15 - x ^{2} - 2 y ^{2}}

Solution steps

\frac{\partial}{\partial x} (15 - x^{2} - 2 y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 15 - x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (15 - x^{2} - 2 y^{2})

= \frac{1}{2 15 - x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (15 - x^{2} - 2 y^{2})

\frac{\partial}{\partial x} (15 - x^{2} - 2 y^{2}) = - 2 x

= \frac{1}{2 15 - x ^{2} - 2 y ^{2}} (- 2 x)

Simplify

\frac{1}{2 15 - x ^{2} - 2 y ^{2}} (- 2 x) : - \frac{x}{15 - x ^{2} - 2 y ^{2}}

= - \frac{x}{15 - x ^{2} - 2 y ^{2}}

(\frac{d y}{d x}) = \frac{( x ^{2} ( y ^{5} - 2 ) )}{y ^{4}}

\frac{d y}{d x} = \frac{5 ( 3 - y ) ^{2}}{( 3 - x ) ^{2}}, y (- 2) = - 5

\int_{0}^{\frac{π}{4}} (sin (4 x)) d x

x^{3} - 12 x^{2} + 20 x, - x^{3} + 12 x^{2} - 20 x

f (x) = 7 tan (x)

What is (\partial)/(\partial x)(sqrt(15-x^2-2y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(15-x^2-2y^2)) is -x/(sqrt(15-x^2-2y^2))