What is (\partial)/(\partial x)(sqrt(10-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(10-x^2-y^2)) is -x/(sqrt(10-x^2-y^2))

(\partial)/(\partial x)(sqrt(10-x^2-y^2))

\frac{\partial}{\partial x} (10 - x^{2} - y^{2})

- \frac{x}{10 - x ^{2} - y ^{2}}

Solution steps

\frac{\partial}{\partial x} (10 - x^{2} - y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 10 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (10 - x^{2} - y^{2})

= \frac{1}{2 10 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (10 - x^{2} - y^{2})

\frac{\partial}{\partial x} (10 - x^{2} - y^{2}) = - 2 x

= \frac{1}{2 10 - x ^{2} - y ^{2}} (- 2 x)

Simplify

\frac{1}{2 10 - x ^{2} - y ^{2}} (- 2 x) : - \frac{x}{10 - x ^{2} - y ^{2}}

= - \frac{x}{10 - x ^{2} - y ^{2}}

\frac{\partial}{\partial x} (ln (8 - x e^{- x}))

\frac{d}{d x} ((\frac{4 x ^{2}}{8 - x})^{3})

x \to 1 lim (\frac{x ^{3} + 1}{x - 1})

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

d er i v a t i v e 3 sin (θ)

What is (\partial)/(\partial x)(sqrt(10-x^2-y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(10-x^2-y^2)) is -x/(sqrt(10-x^2-y^2))