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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\frac{d}{d x} (\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 1 lim (\frac{x + 1}{x + 2})

f^{^{'}} (x) = x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1

\int_{1}^{3} 5 x^{3} ln (x) d x

\int_{0}^{2} \frac{1}{16 - x ^{2}} d x

\frac{d}{d θ} (3 sin (2 θ))

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)