What is (\partial)/(\partial x)(sqrt(62-2x^2-2y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(62-2x^2-2y^2)) is -(2x)/(sqrt(62-2x^2-2y^2))

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(\partial)/(\partial x)(sqrt(62-2x^2-2y^2))

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

- \frac{2 x}{62 - 2 x ^{2} - 2 y ^{2}}

Solution steps

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 62 - 2 x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

= \frac{1}{2 62 - 2 x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2}) = - 4 x

= \frac{1}{2 62 - 2 x ^{2} - 2 y ^{2}} (- 4 x)

Simplify

\frac{1}{2 62 - 2 x ^{2} - 2 y ^{2}} (- 4 x) : - \frac{2 x}{62 - 2 x ^{2} - 2 y ^{2}}

= - \frac{2 x}{62 - 2 x ^{2} - 2 y ^{2}}

\frac{d}{d x} (ln (\frac{1}{1 + x}))

f (x) = arcsec (x)

y^{^{'}} + 7 x e^{y} = 0

\int_{- 1}^{0} 7 x d x

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

What is (\partial)/(\partial x)(sqrt(62-2x^2-2y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(62-2x^2-2y^2)) is -(2x)/(sqrt(62-2x^2-2y^2))