How do you differentiate $y=ln((x-1)/(x^2+1))$ ?

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Answer 1 · 2016-07-27T11:39:47

$dy/dx=(-x^2+2x+1)/((x^2+1)(x-1))$

$y=ln((x-1)/(x^2+1))$

$y=ln(x-1)-ln(x^2+1)$

Use quotient rule of logarithms

Now differentiate

$dy/dx=1/(x-1)-1/(x^2+1) * d/dx(x^2+1)$ Use chain rule

$dy/dx=1/(x-1)-1/(x^2+1)*2x$

$dy/dx=1/(x-1)-(2x)/(x^2+1)$ Take the lcd as ((x-1) (x^2+1)

$dy/dx=((x^2+1)/((x^2+1)(x-1)) )-((2x)(x-1))/((x^2+1)(x-1)) )$
$dy/dx=(x^2+1-2x^2+2x)/((x^2+1)(x-1)$
$dy/dx=(-x^2+2x+1)/((x^2+1)(x-1))$

How do you differentiate y=ln((x-1)/(x^2+1))y=ln((x-1)/(x^2+1))?