What is the derivative of (x^2-2x-3)/((x-1)^2) ?

The derivative of (x^2-2x-3)/((x-1)^2) is 8/((x-1)^3)

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

The first derivative of (x^2-2x-3)/((x-1)^2) is 8/((x-1)^3)

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derivative of

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

\frac{8}{( x - 1 ) ^{3}}

Solution steps

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

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

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

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

Simplify

\frac{( 2 x - 2 ) ( x - 1 ) ^{2} - 2 ( x - 1 ) ( x ^{2} - 2 x - 3 )}{( ( x - 1 ) ^{2} ) ^{2}} : \frac{8}{( x - 1 ) ^{3}}

= \frac{8}{( x - 1 ) ^{3}}

144 y^{^{'''}} + 168 y^{^{''}} + 49 y^{^{'}} = 0

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

x \frac{d y}{d x} = y + x e^{x}

\frac{d}{d x} (\frac{3}{x + 5})

\int 19 tan^{3} (x) d x

What is the derivative of (x^2-2x-3)/((x-1)^2) ?
The derivative of (x^2-2x-3)/((x-1)^2) is 8/((x-1)^3)
What is the first derivative of (x^2-2x-3)/((x-1)^2) ?
The first derivative of (x^2-2x-3)/((x-1)^2) is 8/((x-1)^3)