What is the derivative of (x^2-x-2)/(x^2+x-6) ?

The derivative of (x^2-x-2)/(x^2+x-6) is 2/((x+3)^2)

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

The first derivative of (x^2-x-2)/(x^2+x-6) is 2/((x+3)^2)

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derivative of

\frac{x ^{2} - x - 2}{x ^{2} + x - 6}

\frac{2}{( x + 3 ) ^{2}}

Solution steps

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

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

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

\frac{d}{d x} (x^{2} + x - 6) = 2 x + 1

= \frac{( 2 x - 1 ) ( x ^{2} + x - 6 ) - ( 2 x + 1 ) ( x ^{2} - x - 2 )}{( x ^{2} + x - 6 ) ^{2}}

Simplify

\frac{( 2 x - 1 ) ( x ^{2} + x - 6 ) - ( 2 x + 1 ) ( x ^{2} - x - 2 )}{( x ^{2} + x - 6 ) ^{2}} : \frac{2}{( x + 3 ) ^{2}}

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

y^{^{''}} - 8 y^{^{'}} + 9 y = 0

x \to 4 lim (\frac{x ^{2} - 4}{x - 4})

\frac{\partial}{\partial y} (\frac{x}{y} - \frac{y}{z})

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

x \to - 1 lim (2 x^{2} - 5 x + 1)

What is the derivative of (x^2-x-2)/(x^2+x-6) ?
The derivative of (x^2-x-2)/(x^2+x-6) is 2/((x+3)^2)
What is the first derivative of (x^2-x-2)/(x^2+x-6) ?
The first derivative of (x^2-x-2)/(x^2+x-6) is 2/((x+3)^2)