What is (\partial)/(\partial x)(sqrt(16-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(16-x^2-y^2)) is -x/(sqrt(16-x^2-y^2))

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(\partial)/(\partial x)(sqrt(16-x^2-y^2))

\frac{\partial}{\partial x} (16 - x^{2} - y^{2})

- \frac{x}{16 - x ^{2} - y ^{2}}

Solution steps

\frac{\partial}{\partial x} (16 - x^{2} - y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 16 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (16 - x^{2} - y^{2})

= \frac{1}{2 16 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (16 - x^{2} - y^{2})

\frac{\partial}{\partial x} (16 - x^{2} - y^{2}) = - 2 x

= \frac{1}{2 16 - x ^{2} - y ^{2}} (- 2 x)

Simplify

\frac{1}{2 16 - x ^{2} - y ^{2}} (- 2 x) : - \frac{x}{16 - x ^{2} - y ^{2}}

= - \frac{x}{16 - x ^{2} - y ^{2}}

y = e^{- 2 x} + 1, y = 0, x = 0, x = 3

\frac{\partial}{\partial t} (3 e^{s i n (x + 4 c t)})

f (x) = x^{6 x}

y^{^{'}} = \frac{( 4 x ^{3} + y ^{3} )}{x y ^{2}}

n = 0 \sum \infty \frac{1}{2 n !}

What is (\partial)/(\partial x)(sqrt(16-x^2-y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(16-x^2-y^2)) is -x/(sqrt(16-x^2-y^2))