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

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

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

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

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\frac{d}{d 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}}

\frac{d y}{d x} = 4 e^{3 x - 4 y}

\frac{d}{d x} (\frac{6}{2 ^{x}})

\frac{d}{d x} (\frac{x}{x + 5})

x \frac{d y}{d x} = y (y + 1)

f (x) = \frac{6 x}{x + 2}, (4, 4)

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