What is the derivative of f(x)=(x^2+5x+6)/(x^2-4) ?

The derivative of f(x)=(x^2+5x+6)/(x^2-4) is -5/((x-2)^2)

What is the first derivative of f(x)=(x^2+5x+6)/(x^2-4) ?

The first derivative of f(x)=(x^2+5x+6)/(x^2-4) is -5/((x-2)^2)

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derivative of

f (x) = \frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

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

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 5 x + 6}{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} + 5 x + 6 ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (x^{2} + 5 x + 6) = 2 x + 5

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

= \frac{( 2 x + 5 ) ( x ^{2} - 4 ) - 2 x ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{( 2 x + 5 ) ( x ^{2} - 4 ) - 2 x ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}} : - \frac{5}{( x - 2 ) ^{2}}

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

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

\frac{d}{d x} (- 2 e^{- x} cos (x))

y^{^{'}} + 2 y = 0, y (0) = 1

4 y^{^{''}} + 4 y^{^{'}} + 17 = 0

\frac{d}{d x} (\frac{x + 7}{4 x ^{2} e ^{x}})

What is the derivative of f(x)=(x^2+5x+6)/(x^2-4) ?
The derivative of f(x)=(x^2+5x+6)/(x^2-4) is -5/((x-2)^2)
What is the first derivative of f(x)=(x^2+5x+6)/(x^2-4) ?
The first derivative of f(x)=(x^2+5x+6)/(x^2-4) is -5/((x-2)^2)