What is the derivative of ((x^2-x-2))/((x^2-6x+9)) ?

The derivative of ((x^2-x-2))/((x^2-6x+9)) is -(5x-7)/((x-3)^3)

What is the first derivative of ((x^2-x-2))/((x^2-6x+9)) ?

The first derivative of ((x^2-x-2))/((x^2-6x+9)) is -(5x-7)/((x-3)^3)

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derivative of

\frac{( x ^{2} - x - 2 )}{( x ^{2} - 6 x + 9 )}

- \frac{5 x - 7}{( x - 3 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9})

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} - 6 x + 9 ) - \frac{d}{d x} ( x ^{2} - 6 x + 9 ) ( x ^{2} - x - 2 )}{( x ^{2} - 6 x + 9 ) ^{2}}

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

\frac{d}{d x} (x^{2} - 6 x + 9) = 2 x - 6

= \frac{( 2 x - 1 ) ( x ^{2} - 6 x + 9 ) - ( 2 x - 6 ) ( x ^{2} - x - 2 )}{( x ^{2} - 6 x + 9 ) ^{2}}

Simplify

\frac{( 2 x - 1 ) ( x ^{2} - 6 x + 9 ) - ( 2 x - 6 ) ( x ^{2} - x - 2 )}{( x ^{2} - 6 x + 9 ) ^{2}} : - \frac{5 x - 7}{( x - 3 ) ^{3}}

= - \frac{5 x - 7}{( x - 3 ) ^{3}}

x \to 9 + lim (x^{2} + 5 x + 7)

f (x) = x^{3}, (- 1, - 1)

\frac{d y}{d x} = y x

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x y + 2)

\int_{0}^{2} - 7 x^{3} x^{2} + 4 d x

What is the derivative of ((x^2-x-2))/((x^2-6x+9)) ?
The derivative of ((x^2-x-2))/((x^2-6x+9)) is -(5x-7)/((x-3)^3)
What is the first derivative of ((x^2-x-2))/((x^2-6x+9)) ?
The first derivative of ((x^2-x-2))/((x^2-6x+9)) is -(5x-7)/((x-3)^3)