# Thread: Finding a second derivative and solving an equation

1. ## Finding a second derivative and solving an equation

Hey,

I need some assistance finding what's called a second derivative (I couldn't really understand the prof as he was talking about it), and then solving an equation associated with it.

Equation to solve : $\displaystyle f''(x) = 0$

Given:
$\displaystyle f(x)=\frac{x}{x-1}$

Any help greatly appreciated.

2. differentiate the given expression twice (you may have to use quotient rule)

then equate the result to zero and solve for x

3. Just take the derivative twice over. Like so:

First thing I'm gonna do is rewrite so I don't have to use that icky quotient rule.

$\displaystyle \frac{d}{dx}[\frac{x}{x-1}]=\frac{d}{dx}x(x-1)^{-1}=-x(x-1)^{-2}+(x-1)^{-1}$

Now we do it again

$\displaystyle \frac{d^2}{dx}[x(x-1)^{-1}]$ = $\displaystyle \frac{d}{dx}[-x(x-1)^{-2}+(x-1)^{-1}]=2x(x-1)^{-3}-(x-1)^{-2}-(x-1)^{-2}$

factoring out he least common factor

$\displaystyle =(x-1)^{-3}[2x-(x-1)-(x-1)]=\frac{-4x+2}{(x-1)^3}$

Set the thing to equal zero, and solve.

$\displaystyle \frac{-4x+2}{(x-1)^3}=0\Longleftrightarrow{-4x+2}=0$

you can take it from here.
This is called the second derivative test. Check this out:

Concavity and the Second Derivative Test - HMC Calculus Tutorial