What is the second derivative of $x/(x-1)$ and the first derivative of $2/x$ ?

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Answer 1 · 2015-04-07T01:46:28

Question 1
If $f(x) = (g(x))/(h(x))$ then by the Quotient Rule

$f'(x) = (g'(x)*h(x) - g(x)*h'(x))/((g(x))^2)$

So if $f(x) = x/(x-1)$
then the first derivative
$f'(x) = ((1)(x-1) - (x)(1))/x^2$

$= - 1/x^2 = - x^(-2)$

and the second derivative is
$f''(x) = 2x^-3$

Question 2
If $f(x) = 2/x$ this can be re-written as

$f(x) = 2x^-1$

and using standard procedures for taking the derivative
$f'(x) = -2x^-2$
or, if you prefer
$f'(x) = - 2/x^2$

What is the second derivative of x/(x-1)x/(x-1) and the first derivative of 2/x2/x?