$$ f(x) = x - 2\ln(x^2 + 3) $$
I started by using the chain rule on $x^2 + 3$ which gives me $\frac{2x}{x^2} + 3$.
At this point I tried to multiple $\frac{2x}{x^2} + 3$ by $x - 2$ - is this correct? How can I then find the stationary point?
I understand that you may not wish to give me the answer on a silver platter. Perhaps somebody could kindly provide me with an example in a similar format and the steps involved in solving it.
The first derivative of your function can be found like this:
$$ \begin{eqnarray} f'(x) &=& d(x - 2 \ln(x^2 + 3))/dx \\ &=& d(x)/dx - d(2 \ln(x^2 + 3))/dx \\ &=& 1 - \frac{2 d(x^2 + 3)/dx}{x^2 + 3} \\ &=& 1 - \frac{4x}{x^2 + 3} \\ \end{eqnarray} $$
And the stationary points can be found by solving $f'(x) = 0$:
$$ \begin{eqnarray} 1 - \frac{4x}{x^2 + 3} &=& 0 \\ \frac{4x}{x^2 + 3} &=& 1 \\ x^2 + 3 &=& 4x \\ x^2 - 4x + 3 &=& 0 \\ (x - 1)(x - 3) &=& 0 \\ \end{eqnarray} $$
So you have $x = 1$ and $x = 3$.