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

The derivative of ((x^2-2x+3))/((x-1)^2) is -4/((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 -4/((x-1)^3)

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derivative of

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

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

= - \frac{4}{( x - 1 ) ^{3}}

\frac{d}{d x} (e^{3 x} sin (2 x))

y d x + x^{2} d y = 0

\int (\frac{8 x - 3}{12 x - 4 x ^{2} - 5}) d x

\int \frac{sin ( x )}{1 + 2 cos ( x )} d x

l a pl a ce t r an s f or m cos^{2} (3 t)

What is the derivative of ((x^2-2x+3))/((x-1)^2) ?
The derivative of ((x^2-2x+3))/((x-1)^2) is -4/((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 -4/((x-1)^3)