What is the ((-x^2-4x)/((x^2-x-2)^2))^' ?

The ((-x^2-4x)/((x^2-x-2)^2))^' is -(2(-x^3-6x^2-4))/((x^2-x-2)^3)

What is the first ((-x^2-4x)/((x^2-x-2)^2))^' ?

The first ((-x^2-4x)/((x^2-x-2)^2))^' is -(2(-x^3-6x^2-4))/((x^2-x-2)^3)

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(\frac{- x ^{2} - 4 x}{( x ^{2} - x - 2 ) ^{2}})^{^{'}}

- \frac{2 ( - x ^{3} - 6 x ^{2} - 4 )}{( x ^{2} - x - 2 ) ^{3}}

Solution steps

(\frac{- x ^{2} - 4 x}{( x ^{2} - x - 2 ) ^{2}})^{'}

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

(- x^{2} - 4 x)^{'} = - 2 x - 4

((x^{2} - x - 2)^{2})^{'} = 2 (x^{2} - x - 2) (2 x - 1)

= \frac{( - 2 x - 4 ) ( x ^{2} - x - 2 ) ^{2} - 2 ( x ^{2} - x - 2 ) ( 2 x - 1 ) ( - x ^{2} - 4 x )}{( ( x ^{2} - x - 2 ) ^{2} ) ^{2}}

Simplify

\frac{( - 2 x - 4 ) ( x ^{2} - x - 2 ) ^{2} - 2 ( x ^{2} - x - 2 ) ( 2 x - 1 ) ( - x ^{2} - 4 x )}{( ( x ^{2} - x - 2 ) ^{2} ) ^{2}} : - \frac{2 ( - x ^{3} - 6 x ^{2} - 4 )}{( x ^{2} - x - 2 ) ^{3}}

= - \frac{2 ( - x ^{3} - 6 x ^{2} - 4 )}{( x ^{2} - x - 2 ) ^{3}}

x \to 3 lim (x - 2)

\frac{d}{d x} (C e^{4 x})

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

\int \frac{sin ( 5 x )}{5} d x

\frac{d}{d x} (- \frac{45}{8 ( 1 + x ) ^{\frac{7}{2}}})

What is the ((-x^2-4x)/((x^2-x-2)^2))^' ?
The ((-x^2-4x)/((x^2-x-2)^2))^' is -(2(-x^3-6x^2-4))/((x^2-x-2)^3)
What is the first ((-x^2-4x)/((x^2-x-2)^2))^' ?
The first ((-x^2-4x)/((x^2-x-2)^2))^' is -(2(-x^3-6x^2-4))/((x^2-x-2)^3)