What is (\partial)/(\partial x)((41-x^2-y^2)^{1/2}) ?

The answer to (\partial)/(\partial x)((41-x^2-y^2)^{1/2}) is -x/((41-x^2-y^2)^{1/2)}

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(\partial)/(\partial x)((41-x^2-y^2)^{1/2})

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

- \frac{x}{( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 ( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (41 - x^{2} - y^{2})

= \frac{1}{2 ( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (41 - x^{2} - y^{2})

\frac{\partial}{\partial x} (41 - x^{2} - y^{2}) = - 2 x

= \frac{1}{2 ( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}} (- 2 x)

Simplify

\frac{1}{2 ( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}} (- 2 x) : - \frac{x}{( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}}

= - \frac{x}{( 41 - x ^{2} - y ^{2} ) ^{\frac{1}{2}}}

\int_{0}^{1} x e^{- 4 x} d x

\int \frac{1}{x 25 x ^{2} + 16} d x

x + y \frac{d y}{d x} = 0

\frac{d}{d x} (3 e^{2 x} + 2 e^{2 x} x)

f (x) = x^{2} + 7, 16, at x = 3

What is (\partial)/(\partial x)((41-x^2-y^2)^{1/2}) ?
The answer to (\partial)/(\partial x)((41-x^2-y^2)^{1/2}) is -x/((41-x^2-y^2)^{1/2)}