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

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

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

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

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

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

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) = 2 x

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

Simplify

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

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

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

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

y = 34 (3 p + 1) - \frac{2}{5}, p > 0

f (x) = 4 sin (x) + 3 cos (x), at x = 0

(x + 1)^{^{'}}

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