Stacked with this Problem of Calculus I have been struggling for quite some time with the following problem and I would really appreciate some help: Consider $f(d)=\frac{(1-d)\left(1-d^{(\frac{t}{2}-1)}\right)}{(1-d)\left(1-d^{(\frac{t}{2}-1)}\right)+\left(1-d^{(\frac{s}{4})}\right)^2}$, where $t$ and $s$ are both multiples of 4 and $d \in [0,1)$. I conjecture and trying to show that this function is either always decreasing (if $s>t$), or it is initially decreasing and then increasing (if $t>s$). Up to now I have managed to show the case where $t>s$ and the first case only if $s/2>t$. 
I am afraid of missing something obvious, but it's getting a bit frustrating. Any help would be more than welcome. Thanks a lot in advance.
 A: Note: not a full answer, but something which may lend a hand:
Let $t=4m,\ s=4n,\ x=d,$ and define 
$$g(m,x)=(1-x)(1-x^{2m-1}),\quad h(n,x)=(1-x^n)^2.$$
Then your function is 
$$f(x)=\frac{g(m,x)}{g(m,x)+h(n,x)}=\frac{1}{1+h(n,x)/g(n,x)}.$$
Then $f(x)$ will increase or decrease iff $h/g$ respectively decreases or increases, so to keep things easier we can work instead with $g/h$ which will then increase or decrease with $f.$
To get at the derivative of $g/h$ we first factor out the copies of $(1-x)$ and arrive at
$$(g/h)(x)=\frac{1+x+\cdots+x^{2m-2}}{(1+x+\cdots+x^{n-1})^2}.$$
We now apply the quotient rule, ignoring the squared denominator and the extra factor coming from the fact that the denominator of $g/h$ is already squared (i.e. we factor one of those out). We arrive at the following expression, which has the same sign as the derivative of $g/h$ [and so the same sign as $f'$
$$(1+x+\cdots+x^{n-1})(1+2x+3x^2+ \cdots+(2m-2)x^{2m-3}) \\
-2(1+x+\cdots+x^{2m-2})(1+2x+3x^2+\cdots+(n-1)x^{n-2}).\tag{1}$$
We get right away that at $x=0$ we have $f'<0$ in any case. The derivative is hard to work with otherwise, but at least we can see what is happening at $x=1$, by substituting $1$ for $x$ in the expression $(1)$ and using the formula for the sum of the first $k$ integers twice, and otherwise keeping track of the numbers of terms added. When this is done, we get that expression $(1)$ evaluated at $x=1$ has the value $(m-n)(2m-1)n$. That is, when $m=n$ we find $f$ is horizontal at $1$, when $m>n$ $f$ is increasing at $1$ (in agreement with the part of the question OP said was solved) and when $m<n$ we have $f$ decreasing at $1$.
Of course this is not a complete answer, since $f$ might be decreasing at both $0$ and $1$ but not on the whole interval $[0,1).$ Perhaps one might show any $f$ is concave up, which seems so experimentally. 
