What is the second derivative of the function $f (x) = \frac{x}{x - 1}$ $f(x) = (x) / (x - 1)$ ?

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Answer 1 · 2016-10-06T03:34:18

$\frac{d^{2}}{{d x}^{2}} \frac{x}{x - 1} = \frac{2}{{(x - 1)}^{3}}$ $d^2/(dx^2)x/(x-1)=2/(x-1)^3$

For this problem, we will use the quotient rule:

$d/dx f(x)/g(x) = (g(x)f'(x)-f(x)g'(x))/[g(x)]^2$

We can also make it a little easier by dividing to get

$x/(x-1) = 1+1/(x-1)$

First derivative:

$d/dx(1+1/(x-1))$

$= (d/dx1)+(d/dx((x-1)(d/dx1)-1(d/dx(x-1)))/(x-1)^2)$

$=0+((x-1)(0)-(1)(1))/(x-1)^2$

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

Second derivative:

The second derivative is the derivative of the first derivative.

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

$=-((x-1)^2(d/dx1)-1(d/dx(x-1)^2))/[(x-1)^2]^2$

$=-((x-1)^2(0)-1(2(x-1)))/(x-1)^4$

$=2/(x-1)^3$

We could also have used the power rule $d/dx x^n = nx^(n-1)$ for $n!=1$ :

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

$=> d/dx (1+1/(x-1)) =d/dx(1+(x-1)^(-1))$

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

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

$=2(x-2)^(-3)$

which is the same as the result we obtained above.

What is the second derivative of the function f(x) = (x) / (x - 1)f(x)=xx−1f(x) = (x) / (x - 1)?