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

The derivative of (x^2/(sqrt(x^2-4))) is (x^3-8x)/((x^2-4)sqrt(x^2-4))

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

The first derivative of (x^2/(sqrt(x^2-4))) is (x^3-8x)/((x^2-4)sqrt(x^2-4))

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

\frac{x ^{3} - 8 x}{( x ^{2} - 4 ) x ^{2} - 4}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x^{2}) = 2 x

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

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

Simplify

\frac{2 x x ^{2} - 4 - \frac{x}{x ^{2} - 4} x ^{2}}{( x ^{2} - 4 ) ^{2}} : \frac{x ^{3} - 8 x}{( x ^{2} - 4 ) x ^{2} - 4}

= \frac{x ^{3} - 8 x}{( x ^{2} - 4 ) x ^{2} - 4}

\frac{\partial}{\partial y} (\frac{x + y}{x y - 1})

\int \frac{x ^{2} + 3}{x} d x

2 x (5 - 4 x)^{4}, at x = 1

(- 3 x y^{2} + 5 y) + (- 3 x^{2} y + 5 x) y^{^{'}} = 0

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

What is the derivative of (x^2/(sqrt(x^2-4))) ?
The derivative of (x^2/(sqrt(x^2-4))) is (x^3-8x)/((x^2-4)sqrt(x^2-4))
What is the first derivative of (x^2/(sqrt(x^2-4))) ?
The first derivative of (x^2/(sqrt(x^2-4))) is (x^3-8x)/((x^2-4)sqrt(x^2-4))