What is (\partial)/(\partial x)((x^2+y^2+z^2)^{-0.5}) ?

The answer to (\partial)/(\partial x)((x^2+y^2+z^2)^{-0.5}) is -x/((x^2+y^2+z^2)^{1.5)}

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(\partial)/(\partial x)((x^2+y^2+z^2)^{-0.5})

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- 0.5})

- \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}}

Solution steps

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- 0.5})

Treat

y, z

as constants

Apply the chain rule:

- \frac{0.5}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}} \frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2})

= - \frac{0.5}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}} \frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2})

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

= - \frac{0.5}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}} \cdot 2 x

Simplify

- \frac{0.5}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}} \cdot 2 x : - \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}}

= - \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{1.5}}

\int (sec (x) \cdot tan (x)) d x

y^{^{'}} + k y = 0

f (x) = (x^{5} - x^{3} + 3)^{4}

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 12 y = 0

\int \frac{( x ^{3} - 5 x + 2 )}{x ^{\frac{1}{2}}} d x

What is (\partial)/(\partial x)((x^2+y^2+z^2)^{-0.5}) ?
The answer to (\partial)/(\partial x)((x^2+y^2+z^2)^{-0.5}) is -x/((x^2+y^2+z^2)^{1.5)}