Algebraic issues with the calculation of the second derivative of $(a+be^x)/(ae^x+b)$

Question

I'm trying to work out the 2nd derivative of $\dfrac{a+be^x}{ae^x+b}$

I have $f''=\dfrac{(ae^x+b)^2(b^2-a^2)e^x-2ae^x(ae^x+b)(b^2-a^2)e^x}{(ae^x+b)^4}$

There are so many terms, and I'm seriously confused on how to cancel it down.

The mark scheme says I should expect: $\dfrac{(b^2-a^2)(b-ae^x)e^x}{(ae^x+b)^3}$

How do I get from my working to the answer?

Answer 1

From your computation, we have \begin{align} f''(x)&=\dfrac{(ae^x+b)^2(b^2-a^2)e^x-2ae^x(ae^x+b)(b^2-a^2)e^x}{(ae^x+b)^4}\\ &=\dfrac{(b^2-a^2)(ae^x+b)e^x[(ae^x+b)-2ae^x]}{(ae^x+b)^4}\\ &=\dfrac{(b^2-a^2)e^x(b-ae^x)}{(ae^x+b)^3}. \end{align}

Answer 2

If you are confused by so many terms it might help rewriting the formulas with some substitutions, to clear things up. I suggest setting $b^2-a^2=A$ and $ae^x+b=B(x)$.