What is the derivative of f(x)=(2-5x-x^2)/(x^2-4) ?

The derivative of f(x)=(2-5x-x^2)/(x^2-4) is (5x^2+4x+20)/((x^2-4)^2)

What is the first derivative of f(x)=(2-5x-x^2)/(x^2-4) ?

The first derivative of f(x)=(2-5x-x^2)/(x^2-4) is (5x^2+4x+20)/((x^2-4)^2)

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derivative of

f (x) = \frac{2 - 5 x - x ^{2}}{x ^{2} - 4}

\frac{5 x ^{2} + 4 x + 20}{( x ^{2} - 4 ) ^{2}}

Solution steps

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

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

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

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

Expand

(- 2 x - 5) (x^{2} - 4) - 2 x (2 - 5 x - x^{2}) : 5 x^{2} + 4 x + 20

= \frac{5 x ^{2} + 4 x + 20}{( x ^{2} - 4 ) ^{2}}

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

\int_{0}^{0.6} 2500 x d x

(x^{2} + 4) d y = (2 x - 10 x y) d x

x \to 0 + lim (\frac{( ln ( x ) )}{x})

\frac{d}{d x} (\frac{3}{16 + x ^{4}})

What is the derivative of f(x)=(2-5x-x^2)/(x^2-4) ?
The derivative of f(x)=(2-5x-x^2)/(x^2-4) is (5x^2+4x+20)/((x^2-4)^2)
What is the first derivative of f(x)=(2-5x-x^2)/(x^2-4) ?
The first derivative of f(x)=(2-5x-x^2)/(x^2-4) is (5x^2+4x+20)/((x^2-4)^2)