What is the derivative of 1/(sqrt(x^2+2y^2-5)) ?

The derivative of 1/(sqrt(x^2+2y^2-5)) is -(x+2y\frac{dy)/(dx)}{(x^2+2y^2-5)^{3/2}}

What is the first derivative of 1/(sqrt(x^2+2y^2-5)) ?

The first derivative of 1/(sqrt(x^2+2y^2-5)) is -(x+2y\frac{dy)/(dx)}{(x^2+2y^2-5)^{3/2}}

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

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

Solution steps

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

Apply exponent rule:

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

= \frac{d}{d x} ((x^{2} + 2 y^{2} - 5)^{- \frac{1}{2}})

Apply the chain rule:

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

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

\frac{d}{d x} (x^{2} + 2 y^{2} - 5) = 2 x + 4 y \frac{d y}{d x}

= - \frac{1}{2 ( x ^{2} + 2 y ^{2} - 5 ) ^{\frac{3}{2}}} (2 x + 4 y \frac{d y}{d x})

Simplify

- \frac{1}{2 ( x ^{2} + 2 y ^{2} - 5 ) ^{\frac{3}{2}}} (2 x + 4 y \frac{d y}{d x}) : - \frac{x + 2 y \frac{d y}{d x}}{( x ^{2} + 2 y ^{2} - 5 ) ^{\frac{3}{2}}}

= - \frac{x + 2 y \frac{d y}{d x}}{( x ^{2} + 2 y ^{2} - 5 ) ^{\frac{3}{2}}}

f (x) = 5 x + 6 x

f (x) = 20 x

x \to 0 lim (\frac{sin ( 13 x )}{x})

\int \frac{2900}{1 + 0.25 x} d x

f (x) = \frac{1}{4 x}, at x = 9

What is the derivative of 1/(sqrt(x^2+2y^2-5)) ?
The derivative of 1/(sqrt(x^2+2y^2-5)) is -(x+2y\frac{dy)/(dx)}{(x^2+2y^2-5)^{3/2}}
What is the first derivative of 1/(sqrt(x^2+2y^2-5)) ?
The first derivative of 1/(sqrt(x^2+2y^2-5)) is -(x+2y\frac{dy)/(dx)}{(x^2+2y^2-5)^{3/2}}