What is the derivative of 1/(sqrt(x^2+y^2-4)) ?

The derivative of 1/(sqrt(x^2+y^2-4)) is -(x+y\frac{dy)/(dx)}{(x^2+y^2-4)^{3/2}}

What is the first derivative of 1/(sqrt(x^2+y^2-4)) ?

The first derivative of 1/(sqrt(x^2+y^2-4)) is -(x+y\frac{dy)/(dx)}{(x^2+y^2-4)^{3/2}}

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\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 4})

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

Solution steps

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 4})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

- \frac{1}{2 ( x ^{2} + y ^{2} - 4 ) ^{\frac{3}{2}}} \frac{d}{d x} (x^{2} + y^{2} - 4)

= - \frac{1}{2 ( x ^{2} + y ^{2} - 4 ) ^{\frac{3}{2}}} \frac{d}{d x} (x^{2} + y^{2} - 4)

\frac{d}{d x} (x^{2} + y^{2} - 4) = 2 x + 2 y \frac{d y}{d x}

= - \frac{1}{2 ( x ^{2} + y ^{2} - 4 ) ^{\frac{3}{2}}} (2 x + 2 y \frac{d y}{d x})

Simplify

- \frac{1}{2 ( x ^{2} + y ^{2} - 4 ) ^{\frac{3}{2}}} (2 x + 2 y \frac{d y}{d x}) : - \frac{x + y \frac{d y}{d x}}{( x ^{2} + y ^{2} - 4 ) ^{\frac{3}{2}}}

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

\int_{0}^{x} x cos (a x) d x

\frac{d y}{d x} + y = sin (x)

\int_{- \frac{π}{4}}^{0} 2 sec^{3} (x) d x

\frac{d}{d x} (1 - \frac{x}{2} - ln (2 + 9 x - x^{2}))

\int e^{x} sin (x) sin (3 x) d x

What is the derivative of 1/(sqrt(x^2+y^2-4)) ?
The derivative of 1/(sqrt(x^2+y^2-4)) is -(x+y\frac{dy)/(dx)}{(x^2+y^2-4)^{3/2}}
What is the first derivative of 1/(sqrt(x^2+y^2-4)) ?
The first derivative of 1/(sqrt(x^2+y^2-4)) is -(x+y\frac{dy)/(dx)}{(x^2+y^2-4)^{3/2}}