What is the derivative of (x^2-6x+1/((x^2-1)^2)) ?

The derivative of (x^2-6x+1/((x^2-1)^2)) is (2(-x^3+9x^2-3x+3))/((x^2-1)^3)

What is the first derivative of (x^2-6x+1/((x^2-1)^2)) ?

The first derivative of (x^2-6x+1/((x^2-1)^2)) is (2(-x^3+9x^2-3x+3))/((x^2-1)^3)

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

\frac{2 ( - x ^{3} + 9 x ^{2} - 3 x + 3 )}{( x ^{2} - 1 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{( 2 x - 6 ) ( x ^{2} - 1 ) ^{2} - 4 x ( x ^{2} - 1 ) ( x ^{2} - 6 x + 1 )}{( ( x ^{2} - 1 ) ^{2} ) ^{2}}

Simplify

\frac{( 2 x - 6 ) ( x ^{2} - 1 ) ^{2} - 4 x ( x ^{2} - 1 ) ( x ^{2} - 6 x + 1 )}{( ( x ^{2} - 1 ) ^{2} ) ^{2}} : \frac{2 ( - x ^{3} + 9 x ^{2} - 3 x + 3 )}{( x ^{2} - 1 ) ^{3}}

= \frac{2 ( - x ^{3} + 9 x ^{2} - 3 x + 3 )}{( x ^{2} - 1 ) ^{3}}

\int cos (x) (5 + 5 sin^{2} (x)) d x

\int x^{2} - 9 x + 3 d x

f (x) = x x, (9, 27)

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

What is the derivative of (x^2-6x+1/((x^2-1)^2)) ?
The derivative of (x^2-6x+1/((x^2-1)^2)) is (2(-x^3+9x^2-3x+3))/((x^2-1)^3)
What is the first derivative of (x^2-6x+1/((x^2-1)^2)) ?
The first derivative of (x^2-6x+1/((x^2-1)^2)) is (2(-x^3+9x^2-3x+3))/((x^2-1)^3)