What is the derivative of f(x)=(2-3x-x^2)/(x^2-4) ?

The derivative of f(x)=(2-3x-x^2)/(x^2-4) is (3x^2+4x+12)/((x^2-4)^2)

What is the first derivative of f(x)=(2-3x-x^2)/(x^2-4) ?

The first derivative of f(x)=(2-3x-x^2)/(x^2-4) is (3x^2+4x+12)/((x^2-4)^2)

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derivative of

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

\frac{3 x ^{2} + 4 x + 12}{( x ^{2} - 4 ) ^{2}}

Solution steps

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

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

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

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

Expand

(- 2 x - 3) (x^{2} - 4) - 2 x (2 - 3 x - x^{2}) : 3 x^{2} + 4 x + 12

= \frac{3 x ^{2} + 4 x + 12}{( x ^{2} - 4 ) ^{2}}

\int \frac{4 x}{x ^{2} - 1} d x

y^{^{''}} + 4 y^{^{'}} = 2 x - 3 e^{x}

\int 5 cos^{3} (x) d x

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

\frac{d}{d x} (\frac{1 + x}{1 - x})

What is the derivative of f(x)=(2-3x-x^2)/(x^2-4) ?
The derivative of f(x)=(2-3x-x^2)/(x^2-4) is (3x^2+4x+12)/((x^2-4)^2)
What is the first derivative of f(x)=(2-3x-x^2)/(x^2-4) ?
The first derivative of f(x)=(2-3x-x^2)/(x^2-4) is (3x^2+4x+12)/((x^2-4)^2)